The Lebesgue Integral

Abstract

Let $\{X,\mathcal {A},\mu \}$ be a measure space and $E\in \mathcal {A}$.

Appendices

Problems and Complements

1c Measurable Functions

In the problems 1.1–1.12 $\{X,\mathcal {A},\mu \}$ is a measure space and $E\in \mathcal {A}$.

1.1. :

The characteristic function of a set E is measurable if and only if E is measurable.

1.2. :

A function f is measurable if and only if its restriction to any measurable subset of its domain is measurable.

1.3. :

A function f is measurable if and only if $f^+$ and $f^-$ are both measurable.

1.4. :

Let $\{X,\mathcal {A},\mu \}$ be complete. A function defined on a set of measure zero is measurable.

1.5. :

Let $\{X,\mathcal {A},\mu \}$ not be complete. Then there exists a measurable set A of measure zero that contains a nonmeasurable set B. The two functions $f=\chi _A$ and $g=\chi _{A-B}$ differ on a set of outer measure zero. However f is measurable and g is not.

1.6. :

If f is measurable then $[f=c]$ is measurable for all c in the range of f. The converse is false.

1.7. :

Let f be measurable. Then also $|f|^{p-1}f$ is measurable for all $p>0$.

1.8. :

|f| measurable does not imply that f is measurable. Likewise $f^2$ measurable does not imply that f is measurable.

1.9. :

A function f is measurable if and only if its preimage of a Borel set is measurable.

1.10. :

Let $\{f_n\}$ be a sequence of measurable functions from E into $\mathbb {R}^*$. The subset of E where $\lim f_n$ exists is measurable.

1.11. :

The supremum(infimum) of an uncountable family $\{f_\alpha \}$ of measurable functions, need not be measurable.

1.12. :

Upper(lower) semi-continuous functions $f:E\rightarrow \mathbb {R}^*$ are measurable.

In the problems 1.13–1.19, $\{\mathbb {R}^N,\mathcal {M},\mu \}$ is $\mathbb {R}^N$ with the Lebesgue measure and $E\in \mathcal {M}$.  

1.13. :

There exists a nonmeasurable function $f:\mathbb {R}\rightarrow \mathbb {R}$ such that $f^{-1}(y)$ is measurable for all $y\in \mathbb {R}$. Hint: given a nonmeasurable set $E\subset \mathbb {R}$, define $f(x)=x$ for $x\in \mathbb {R}-E$ and $f(x)=-x$ for $x\in E$.

1.14. :

A monotone function f in some interval $(a,b)\subset \mathbb {R}$ is measurable.

1.15. :

Let $f:\mathbb {R}\rightarrow \mathbb {R}$ be measurable and let $g:\mathbb {R}\rightarrow \mathbb {R}$ be continuous. The composition $g(f):E\rightarrow \mathbb {R}$ is measurable. However the composition $f(g):E\rightarrow \mathbb {R}$ in general is not measurable. To construct a counterexample consider the function of § 14 of Chap. 3.

1.16. :

Let $f:[0,1]\rightarrow \mathbb {R}^*$ be measurable. Then $\chi _{\mathbf {Q}\cap [0,1]}(f)$, is measurable.

1.17. :

Let $f:\mathbb {R}^2\rightarrow \mathbb {R}^*$ be such that $f(\cdot ,y)$ is continuous for all $y\in \mathbb {R}$ and $f(x,\cdot )$ is measurable for all $x\in \mathbb {R}$. Prove that f is measurable.

We indicate two approaches to this statement. The first is based on the following lemma.

 

Lemma 1.1c

Let $c\in \mathbb {R}$ and let $y\in \mathbb {R}$ be fixed. Then $f(x,y)\ge c$ if and only if, for all $n\in \mathbb {N}$ there exists a rational number $r_m$ such that

$$\begin{aligned} |x-r_m|<{\textstyle \frac{1}{n}}\qquad \text { and }\qquad f(r_m,y)>c-{\textstyle \frac{1}{n}}. \end{aligned}$$

Proof

Since $g=f(\cdot ,y)$ is continuous $g^{-1}(c-\frac{1}{n},\infty )$ is open and we may select $r_m\in \mathbb {Q}$ with the indicated properties. Conversely, let x satisfy the property of the lemma, and assume by contradiction that $g(x)<c$. Since g is continuous, there exists $k\in \mathbb {N}$ such that $g(x)<c-\frac{1}{k}$. Since $x\in g^{-1}(-\infty ,c-\frac{1}{k})$, and since this set is open, there exists $h\in \mathbb {N}$ such that

$$\begin{aligned} (x-{\textstyle \frac{1}{h}},x+{\textstyle \frac{1}{h}}) \subset g^{-1}(-\infty ,c-{\textstyle \frac{1}{k}}). \end{aligned}$$

Take $n=\max \{k;h\}$ and select any $r\in \mathbb {Q}$ such that

$$\begin{aligned} |x-r|<{\textstyle \frac{1}{m}}\quad \text { and }\quad r\in g^{-1}(-\infty ,c-{\textstyle \frac{1}{k}}). \end{aligned}$$

For such choices

$$\begin{aligned} g(r)=f(r,y)<c-{\textstyle \frac{1}{n}}\le c-{\textstyle \frac{1}{k}}. \end{aligned}$$

$\blacksquare$

Using the lemma

$$\begin{aligned}{}[f\ge c]=\mathop {\textstyle {\bigcap }}\limits _n\mathop {\textstyle {\bigcup }}\limits _m\Big ( \big [f(r_m,\cdot )>c-{\textstyle \frac{1}{m}}\big ]\times B_{\frac{1}{n}}(r_m)\Big ). \end{aligned}$$

The second approach is based on the following construction. For $n\in \mathbb {N}$ and $j\in \mathbb {Z}$ set $a_i=i/n$ and

$$\begin{aligned} f_n(x,\cdot )=\frac{f(a_{j+1},\cdot )(x-a_j) -f(a_j,\cdot )(x-a_{j+1})}{a_{j+1}-a_j},\> \quad \text { for }\> a_{j+1}\le x\le a_j. \end{aligned}$$

 

1.18. :

Find an example of a function $f:\mathbb {R}^2\rightarrow \mathbb {R}^*$ measurable in each of its variables separately, and not measurable. Hint: see the Sierpinsky example 13.4. of § 13c.

1.19. :

Let T be a linear nonsingular transformation of $\mathbb {R}^N$ onto itself and let $f:\mathbb {R}^N\rightarrow \mathbb {R}^*$ be measurable. Prove that $f(T):\mathbb {R}^N\rightarrow \mathbb {R}^*$ is measurable. Hint: By § 12.3c of Chap. 3 it suffices to show that

$$\begin{aligned}{}[f(T)>c]=T^{-1}[f>c]\quad \text { for all }\> c\in \mathbb {R}. \end{aligned}$$

 

1.1 1.1c Sublevel Sets

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $f:X\rightarrow \mathbb {R}$ be measurable. For $t\in \mathbb {R}$, the sublevel set of f at t, is defined as $E_t=[f\le t]$. By the definition

$$\begin{aligned} E_s\subset E_t \>\text { for }\> s<t;\quad \mathop {\textstyle {\bigcup }}\limits E_t=X;\quad \mathop {\textstyle {\bigcap }}\limits E_t=\emptyset ;\quad \mathop {\textstyle {\bigcap }}\limits _{s<t} E_t=E_s. \end{aligned}$$

(1.1c)

Conversely, given a collection of measurable sets $\{E_t\}_{t\in \mathbb {R}}$ in $\varOmega$, satisfying (1.1c) there exists a measurable function $f:X\rightarrow \mathbb {R}$ for which $E_t=[f\le t]$. Verify that (von Neumann [114])

$$\begin{aligned} X\ni x\rightarrow f(x)=\inf \{t:x\in E_t\} \end{aligned}$$

is such a function and prove that it is unique.

2c The Egorov–Severini Theorem

 

2.1. :

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $E\in \mathcal {A}$ be of finite measure. Let $\{f_n\}$ be a sequence of measurable functions from E into $\mathbb {R}^*$. Assume that for a.e. $x\in E$ the set $\{f_n(x)\}$ is bounded. For every $\varepsilon >0$ there exists a measurable set $E_\varepsilon \subset E$ and a positive number $k_\varepsilon$ such that

$$\begin{aligned} \mu (E-E_\varepsilon )\le \varepsilon \quad \text { and }\> |f_n|\le k_\varepsilon \quad \text { on }\> E_\varepsilon \quad \text { for all }\> n\in \mathbb {N}. \end{aligned}$$

2.2. :

Let $\mu$ be the counting measure on $2^{\mathbb {N}}$. Define $f_n$ as the characteristic function of $\{1,\dots ,n\}$. The sequence $\{f_n\}$ converges to 1 everywhere in $\mathbb {N}$ but not in measure.

 

Proposition 2.1c

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $E\in \mathcal {A}$. Let $\{f_n\}$ and f be measurable functions from E into $\mathbb {R}^*$. Assume that f is finite a.e. in E. Then $\{f_n\}\rightarrow f$ a.e. in E if and only if for all $\eta >0$

$$\begin{aligned} \lim \mu \big (\mathop {\textstyle {\bigcup }}\limits _{j=n}^\infty \left[ |f_j-f|\ge \eta \right] \big )=0. \end{aligned}$$

(2.1c)

Hint: Denoting by A the set where $\{f_n\}$ is not convergent

$$\begin{aligned} A=\mathop {\textstyle {\bigcup }}\limits _m\limsup \big [|f_n-f|\ge {\textstyle \frac{1}{m}}\big ]. \end{aligned}$$

Then $\mu (A)=0$ if and only if (2.1c) holds.

3c Approximating Measurable Functions by Simple Functions

Proposition 3.1c

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $f:X\rightarrow \mathbb {R}^*$ be nonnegative and measurable. There exists a countable collection $\{E_j\}\subset \mathcal {A}$ such that $X=\mathop {\textstyle {\bigcup }}\limits E_j$ and

$$\begin{aligned} f=\mathop {\textstyle {\sum }}\limits _{j\in \mathbb {N}} \frac{1}{j}\chi _{E_j}. \end{aligned}$$

(3.1c)

Proof

Let $E_1=[f\ge 1]$ and $f_1=\chi _{E_1}$. Then, for $j\ge 2$ define recursively

$$\begin{aligned} E_j=\Big [f\ge \frac{1}{j}+f_{j-1}\Big ]\qquad \text { where }\qquad f_j=\mathop {\textstyle {\sum }}\limits _{i=1}^{j}\frac{1}{i}\chi _{E_i}. \end{aligned}$$

By construction $f\ge f_j$ for all j. If $f(x)=\infty$ then $x\in E_j$ for all j. Hence f(x) has the representation (3.1c). If $f(x)<\infty$ then $x\notin E_j$ for infinitely many j. For such an x

$$\begin{aligned} 0\le f(x)-f_j(x)\le \frac{1}{j+1}\quad \text { for infinitely many}\, j. \end{aligned}$$

$\blacksquare$

The functions $f_j$ are simple and hence the proposition gives an alternative proof of Proposition 3.1. However simple functions, even in canonical form, cannot be written, in general, as a finite sum of the type of (3.1c). As an example:  

3.1. :

Find the representation (3.1c) of a positive, real multiple of the characteristic function of the interval (0, 1).

3.2. :

Find the representation (3.1c) of a simple function taking only two positive values on distinct, measurable sets.

 

4c Convergence in Measure

In the problems 4.1–4.7, $\{X,\mathcal {A},\mu \}$ is complete measure space, and $E\in \mathcal {A}$ is of finite measure.  

4.1. :

Let $f:E\rightarrow \mathbb {R}^*$ be measurable and assume that $|f|>0$ a.e. on E. For every $\varepsilon >0$ there exists a measurable set $E_\varepsilon \subset E$ and a positive number $\delta _\varepsilon$ such that $\mu (E-E_\varepsilon )\le \varepsilon$ and $|f|>\delta _\varepsilon$ on $E_\varepsilon$.

4.2. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of measurable functions converging to f a.e. in E. Assume that $|f|>0$ and $|f_n|>0$ a.e. on E for all $n\in \mathbb {N}$. For every $\varepsilon >0$ there exists a measurable set $E_\varepsilon \subset E$ and a positive number $\delta _\varepsilon$ such that

$$\begin{aligned} \mu (E-E_\varepsilon )\le \varepsilon \quad \text { and }\> |f_n|>\delta _\varepsilon \quad \text { on }\>E_\varepsilon \quad \text { for all }\>n\in \mathbb {N}. \end{aligned}$$

4.3. :

Let $\mu$ be the counting measure on the rationals of [0, 1]. Then convergence in measure is equivalent to uniform convergence .

4.4. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of measurable and a.e. finite functions. There exists a sequence of positive numbers $\{k_n\}$ such that $f_nk_n^{-1}\rightarrow 0$ a.e. in E.

4.5. :

Let $\{f_n\},\,\{g_n\}: E\rightarrow \mathbb {R}^*$ be sequences of measurable functions converging in measure to f and g, respectively, and let $\alpha ,\beta \in \mathbb {R}$. Then

$$\begin{aligned} \{\alpha f_n+\beta g_n\},\quad \{|f_n|\},\quad \{f_n g_n\}\quad \longrightarrow \quad \alpha f+\beta g,\quad |f|,\quad fg \end{aligned}$$

in measure. Moreover if $f\ne 0$ a.e. on E and $f_n\ne 0$ a.e. on E for all n, then $1/f_n$ converges to 1 / f in measure. (Hint: Use 4.1–4.2).

4.6. :

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $\{E_n\}$ be a sequence of measurable sets. The sequence $\{\chi _{E_n}\}$ converges in measure if and only if $d(E_n;E_m)\rightarrow 0$ as $n,m\rightarrow \infty$ (see 2.2. and 3.7. of the Complements of Chap. 3).

4.7. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of measurable and a.e. finite functions. Prove that $\{f_n\}\rightarrow f$ in measure if and only if every subsequence of $\{f_n\}$ contains in turn a subsequence converging to f in measure.

 

7c The Lebesgue Integral of Nonnegative Measurable Functions

1.1 7.1c Comparing the Lebesgue Integral with the PeanoJordan Integral

Let $E\subset \mathbb {R}^N$ be bounded and PeanoJordan measurable, and denote by $\mathcal {P}=\{E_n\}$ a finite partition of E into pairwise disjoint PeanoJordan measurable sets. For a bounded function $f:E\rightarrow \mathbb {R}$ set

$$\begin{aligned} \begin{array}{cc} h_n={\inf }_{x\in E_n}f(x)\qquad &{}\mathcal {F}^-_{\mathcal {P}}=\mathop {\textstyle {\sum }}\limits \,h_n \mu _{{}_{\mathcal {P}-\mathcal {J}}}(E_n)\\ k_n={\sup }_{x\in E_n}f(x)\qquad &{}\mathcal {F}^+_{\mathcal {P}}=\mathop {\textstyle {\sum }}\limits \,k_n \mu _{{}_{\mathcal {P}-\mathcal {J}}}(E_n). \end{array} \end{aligned}$$

A bounded function $f:E\rightarrow \mathbb {R}$ is PeanoJordan integrable if for every $\varepsilon >0$, there exists a partition $\mathcal {P}_\varepsilon$ of E into PeanoJordan measurable sets $E_n$, such that

$$\begin{aligned} \mathcal {F}^+_{\mathcal {P}}-\mathcal {F}^-_{\mathcal {P}}\le \varepsilon . \end{aligned}$$

The sets $E_n$ are also Lebesgue measurable. Therefore,

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits \,h_n\mu _{\mathcal {P}-\mathcal {J}}(E_n)\le \int _Efd\mu \le \mathop {\textstyle {\sum }}\limits \, k_n\mu _{\mathcal {P}-\mathcal {J}}(E_n). \end{aligned}$$

Thus if a bounded function f is PeanoJordan integrable it is also Lebesgue integrable. The converse is false. Indeed the characteristic function of the rationals $\mathbb {Q}$ of [0, 1] is Lebesgue integrable and not PeanoJordan integrable.

This is not longer the case however if f is not bounded. Following Riemann’s notion of improper integral, the function

$$\begin{aligned} f(x)=\frac{1}{x}\,\sin \frac{1}{x}\qquad \text { for }\quad x\in (0,1] \end{aligned}$$

is PeanoJordan integrable in (0, 1) but not Lebesgue integrable.

In the problems 7.1–7.7, $\{X,\mathcal {A},\mu \}$ is a measure space and $E\in \mathcal {A}$.  

7.1. :

Let $f:E\rightarrow \mathbb {R}^*$ be measurable and nonnegative. If $f=\infty$ in a set $\mathcal {E}\subset E$ of measure zero

$$\begin{aligned} \int _Ef d\mu =\int _{E-\mathcal {E}}fd\mu . \end{aligned}$$

7.2. :

Let $f:X\rightarrow \mathbb {R}^*$ be measurable and nonnegative. Then

$$\begin{aligned} \int _Efd\mu =0\quad \text { for all }\> E\in \mathcal {A} \quad \text { implies }\quad f=0\> \text { a.e. in }\> X. \end{aligned}$$

7.3. :

Let $f:E\rightarrow \mathbb {R}^*$ be integrable. If the integral of f over every measurable subset $A\subset E$ is nonnegative, then $f\ge 0$ a.e. on E.

7.4. :

Let $f:\mathbb {R}\rightarrow \mathbb {R}$ be Lebesgue integrable. Then for every $h\in \mathbb {R}$ and every interval $[a,b]\subset \mathbb {R}$

$$\begin{aligned} \int _{[a,b]}fd\mu =\int _{[a+h,b+h]}f(x-h)d\mu . \end{aligned}$$

7.5. :

Construct the Lebesgue integral of a nonnegative $\mu$-measurable function $f:\mathbb {R}^N\rightarrow \mathbb {R}$, when $\mu$ is the Dirac delta measure $\delta _x$ concentrated at some $x\in \mathbb {R}^N$.

7.6. :

Let E be of finite measure. A measurable function $f:E\rightarrow \mathbb {R}^*$ is integrable if and only if $\mathop {\textstyle {\sum }}\limits \mu ([|f|\ge n])<\infty$. Hint:

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits \mu ([|f|\ge n])=\mathop {\textstyle {\sum }}\limits n\mu ([n\le |f|<n+1]). \end{aligned}$$

7.7. :

Let $\{X,\mathcal {A},\mu \}$ be $\mathbb {R}^N$ with the Lebesgue measure. Let T be a linear nonsingular transformation of $\mathbb {R}^N$ onto itself and let $f:\mathbb {R}^N\rightarrow \mathbb {R}^*$ be Lebesgue integrable. Prove that

$$\begin{aligned} \int _Ef(x) d\mu =\frac{1}{|\det T|}\int _{TE} f(y) d\mu . \end{aligned}$$

Hint: § 12.3.1c of the Complements of Chap. 3 and Problem 1.19.

 

1.2 7.2c On the Definition of the Lebesgue Integral

The original definition of Lebesgue was based only on the Lebesgue measure in $\mathbb {R}^N$. The integral of a measurable, nonnegative function f was defined as in (7.1), where the supremum was taken over the class of simple functions $\zeta \le f$, and vanishing outside a set of finite measure. Denote by $\varPhi _f$ such a class of simple functions and observe that

$$\begin{aligned} \int _Efd\mu =\sup _{\varphi \in \varPhi _f}\int _E\varphi d\mu . \end{aligned}$$

Such a definition, while adequate for the Lebesgue measure in $\mathbb {R}^N$, is not adequate for a general measure space $\{X,\mathcal {A},\mu \}$. For example let $\{X,\mathcal {A},\mu \}$ be the measure space of 3.2 of Chap. 3. Then

$$\begin{aligned} \infty =\int _X 1d\mu =\sup _{\varphi \in \varPhi _1}\, \int _X\varphi d\mu =0. \end{aligned}$$

9c More on the Lebesgue Integral

Let $\{X,\mathcal {A},\mu \}$ be a measure space with $\mu (X)=1$ and let $f:X\rightarrow \mathbb {R}$ be $\mu$-measurable. In particular, for all Borel sets $B\in \mathbb {R}$, the set $f^{-1}(B)$ is $\mu$-measurable. Set

$$\begin{aligned} \mu _f(B)=\mu ([f^{-1}(B)])\quad \text { for all Borel sets }\> B\subset \mathbb {R}. \end{aligned}$$

(9.1c)

 

9.1. :

Distribution Measure of a Measurable Function: Verify that $\mu _f$ is a measure on $\mathcal {B}$. Verify that if $f=\chi _E$ for some $E\in \mathcal {A}$ , then for a Borel set $B\subset \mathbb {R}$

$$\begin{aligned} \mu _f(B)=\chi _B(1)\mu (E)+\chi _B(0)\mu (X-E). \end{aligned}$$

9.2. :

A function $h:\mathbb {R}\rightarrow \mathbb {R}^*$ is Borel measurable if the preimage of a Borel set in $\mathbb {R}$ is a Borel set in $\mathbb {R}$. Let h be Borel measurable and integrable with respect to $\mu _f$. Prove that h(f) is integrable in $\{X,\mathcal {A},\mu \}$ and

$$\begin{aligned} \int _{X} h(f) d\mu =\int _{\mathbb {R}} h(t)d\mu _f. \end{aligned}$$

(9.2c)

Hint: Given $\mu _f$, compute the integral of both sides by assuming first that $h=\chi _B$ where $B\subset \mathbb {R}$ is a Borel set. Then in view of (9.1c), the integrals equal $\mu _f(B)=\mu \{f^{-1}(B)\}$. Next assume that h is nonnegative. Compute both sides of (9.2c) by using the definition (7.1) and show that both sides are equal. The general case follows by linear combinations and limiting processes.

9.3. :

For a $\mu$-measurable function f on $\{X,\mathcal {A},\mu \}$ prove that

$$\begin{aligned} \int _X |f|^p d\mu =\int _0^\infty t^p d\mu _f \end{aligned}$$

9.4. :

Equidistributed Measurable Functions: Two measurable function $f_1$, $f_2:X\rightarrow \mathbb {R}$ are equidistributed if

$$\begin{aligned} \mu [f_1^{-1}(B)]=\mu [f_2^{-1}(B)]\quad \text { for every Borel set }\> B\subset \mathbb {R}. \end{aligned}$$

Give an example of equidistributed measurable functions $f_1\ne f_2$. Prove that if $f_1$ and $f_2$ are equi-distributed, then

$$\begin{aligned} \int _X h(f_1) d\mu =\int _X h(f_2)d\mu \quad \text { for all}\, h(\cdot )\, \text {as in}\,\mathbf{9.2}. \end{aligned}$$

9.5. :

Expectation and Variance of a Measurable Function: The expectation E (f) and the variance $\sigma ^2(f)$ of a measurable function f are defined as

$$\begin{aligned} \begin{aligned} E(f)&=\int _X f d\mu =\int _{\mathbb {R}} t d\mu _f\\ \sigma ^2(f)&=\int _X (f-E(f))^2d\mu =\int _{\mathbb {R}}(t-E(f))^2 d\mu _f. \end{aligned} \end{aligned}$$

(9.3c)

The quantity $\sigma (f)$ , is the standard deviation of f. Prove that $\sigma ^2(f)=0$ if and only if there exists $t\in \mathbb {R}$ such that $f=1$ a.e. in X and $t=E(f)$.

9.6. :

Let $f_1$ and $f_2$ be equidistributed measurable functions on $\{X,\mathcal {A},\mu \}$. Then $\sigma ^2(f_1)=\sigma ^2(f_2)$.

 

10c Convergence Theorems

In the problems 10.1–10.10, $\{X,\mathcal {A},\mu \}$ is a measure space and $E\in \mathcal {A}$.  

10.1. :

Let $f:E\rightarrow \mathbb {R}^*$ be integrable and let $\{E_n\}$ be a countable collection of measurable, disjoint subsets of E such that $E=\cup E_n$. Then

$$\begin{aligned} \int _Efd\mu =\mathop {\textstyle {\sum }}\limits \,\int _{E_n}fd\mu . \end{aligned}$$

10.2. :

Let $\mu$ be the counting measure on the positive rationals $\{r_1,r_2,\dots \}$, and let

$$\begin{aligned} f_n(r_j)=\left\{ \begin{array}{ll} {\displaystyle \frac{1}{j}} &{} \qquad \text { if }\> j<n\\ {}\\ 0&{}\qquad \text { if }\> j\ge n. \end{array} \right. \end{aligned}$$

The sequence $\{f_n\}$ converges uniformly to a nonintegrable function.

10.3. :

Let $\mu$ be a finite measure and let $\{f_n\}$ be a sequence of integrable functions converging uniformly in X. The limiting function f is integrable and one can pass to the limit under integral.

 

1.1 10.1c Another Version of Dominated Convergence

Theorem 10.1c

Let $\{f_n\}$ be a sequence of integrable functions in E converging a.e. in E to some f. Assume that there exists a sequence of integrable functions $\{g_n\}$ converging a.e. in E to an integrable function g and such that

$$\begin{aligned} \lim \int _Eg_n d\mu =\int _Eg d\mu . \end{aligned}$$

Assume moreover that $|f_n|\le g_n$ a.e. in E for all $n\in \mathbb {N}$. Then f is integrable and

$$\begin{aligned} \lim \int _Ef_n d\mu =\int _Efd\mu . \end{aligned}$$

 

10.4. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of integrable functions converging to an integrable function f a.e. in E. Then

$$\begin{aligned} \lim \int _E|f_n-f|d\mu =0\quad \text { if and only if }\quad \lim \int _E|f_n|d\mu =\int _E|f|d\mu . \end{aligned}$$

10.5. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of measurable functions satisfying

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits \int _E|f_n|d\mu <\infty . \end{aligned}$$

Then $\mathop {\textstyle {\sum }}\limits f_n$ defines, a.e. in E and integrable function and

$$\begin{aligned} \int _E\mathop {\textstyle {\sum }}\limits f_n d\mu =\mathop {\textstyle {\sum }}\limits \int _Ef_n d\mu . \end{aligned}$$

10.6. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of measurable functions satisfying $|f_n|\le |f|$ for an integrable function $f:E\rightarrow \mathbb {R}$. Then

$$\begin{aligned} \int _E\liminf f_n d\mu \le \liminf \int _Ef_n d\mu \le \limsup \int _Ef_n d\mu \le \int _E\limsup f_n d\mu . \end{aligned}$$

10.7. :

Let E be of finite measure and let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of measurable functions satisfying $|f_n|\le |g|$ for an integrable function $g:E\rightarrow \mathbb {R}$. If $\{f_n\}\rightarrow f$ in measure, then

$$\begin{aligned} \int _Ef d\mu =\lim \int _Ef_n d\mu \quad \text { and }\quad \lim \int _E|f_n-f|d\mu =0. \end{aligned}$$

10.8. :

Let E be of finite measure and let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a sequence of nonnegative measurable functions converging in measure to f. Then

$$\begin{aligned} \int _Ef d\mu \le \liminf \int _Ef_n d\mu . \end{aligned}$$

10.9. :

Prove that in the Egorov-Severini Theorem 2.1 the assumption $\mu (E)<\infty$ can be replaced by $|f_n|\le g$ for an integrable function $g:E\rightarrow \mathbb {R}^*$.

10.10. :

Let $\{f_n\}:E\rightarrow \mathbb {R}^*$ be a nonincreasing sequence of nonnegative, measurable functions. Give an example to show that in general

$$\begin{aligned} \lim \int _Ef_n d\mu \ne \int _E\lim f_n d\mu . \end{aligned}$$

Thus in the Monotone Convergence Theorem 8.1, the monotonicity assumption $f_n\le f_{n+1}$, cannot be replaced with the monotonicity assumption $f_n\ge f_{n+1}$.

 

In the problems 10.11–10.16, $\{X,\mathcal {A},\mu \}$ is $\mathbb {R}^N$ with the Lebesgue measure.  

10.11. :

Let sequences $\{f_n\}$ be defined by

$$\begin{aligned} f_n(x)=\left\{ \begin{array}{ll} n\>&{}\text { if }\> x\in [0,\frac{1}{n}]\\ 0&{}\text { otherwise}, \end{array} \right. \qquad \quad f_n=\left\{ \begin{array}{ll} \frac{1}{n}\>&{}\text { for }\>0\le x\le n\\ 0&{}\text { for }\> x> n. \end{array} \right. \end{aligned}$$

In either case

$$\begin{aligned} 1=\lim \int _{\mathbb {R}^+}f_ndx \ne \int _{\mathbb {R}^+}\lim f_n dx=0. \end{aligned}$$

10.12. :

Let $f:\mathbb {R}\rightarrow \mathbb {R}$ be Lebesgue measurable, nonnegative, and locally bounded. Assume that f is Riemann-integrable on $\mathbb {R}$. Then f is Lebesgue integrable on $\mathbb {R}$ and

$$\begin{aligned} \int _{\mathbb {R}}fd\mu =\lim \int _{-n}^n f d\mu . \end{aligned}$$

10.13. :

Let $\{f_n\}$ be the sequence of nonnegative integrable functions defined on $\mathbb {R}^N$ by

$$\begin{aligned} f_n(x)=\left\{ \begin{array}{ll} {\displaystyle n^N\,\exp {\left\{ \frac{-1}{1-n^2|x|^2}\right\} }} &{} \qquad \text { if }\> |x|<\frac{1}{n}\\ 0&{}\qquad \text { if }\> |x|\ge \frac{1}{n}. \end{array} \right. \end{aligned}$$

The sequence $\{f_n\}$ converges to zero a.e. in $\mathbb {R}^N$ and each $f_n$ is integrable with uniformly bounded integral. However

$$\begin{aligned} \lim \int _{\mathbb {R}^N}f_n dx\ne \int _{\mathbb {R}^N}\lim f_n dx. \end{aligned}$$

10.14. :

Let $\{f_n\}$ be the sequence defined in 7.7 of the Complements of Chap. 2. The assumptions of the dominated convergence theorem fail and

$$\begin{aligned} \lim \int _0^1 f_n(x)dx=\frac{1}{2}\quad \text { and }\quad \int _0^1\lim f_n(x)dx=0. \end{aligned}$$

10.15. :

The two sequences $\{x^n\}$ and $\{n x^n\}$ converge to zero in (0, 1). For the first one can pass to the limit under integral and for the second one cannot.

10.16. :

Integrability and Boundedness: There exist positive, Lebesgue integrable functions on $\mathbb {R}$, that are infinity at all the rationals. Let $\{r_n\}$ be the sequence of the rationals in $\mathbb {R}^+$ and set

$$\begin{aligned} f(x)=\mathop {\textstyle {\sum }}\limits \frac{1}{2^{n}}\frac{e^{-x}}{\sqrt{|x-r_n|}}. \end{aligned}$$

Prove that f is integrable over $\mathbb {R}^+$ (Proposition 10.2).

 

11c Absolute Continuity of the Integral

The proof of Theorem 11.1 shows that for an integrable function f, given $\varepsilon >0$, the corresponding $\delta$ claimed by the theorem, depends upon $\varepsilon$ and f.

A collection $\varPhi$ of integrable functions $f:E\rightarrow \mathbb {R}^*$ is uniformly integrable , if for all $\varepsilon >0$ there exists $\delta =\delta (\varepsilon )>0$ such that the conclusion of Theorem 11.1 holds for all $f\in \varPhi$, for all measurable sets $\mathcal {E}\subset E$ such that $\mu (\mathcal {E})<\delta$ (Vitali [169]).  

11.1. :

If $\varPhi$ is finite then it is uniformly integrable.

11.2. :

Let $\{f_n\}$ be a sequence of integrable functions such that $|f_n|\le g$ for an integrable function g. Then $\{f_n\}$ is uniformly integrable .

11.3. :

Let E be of finite measure and let $\{f_n\}$ be a sequence of uniformly integrable functions converging a.e. in E to an integrable function f . Prove that (Hint: Egorov–Severini theorem)

$$\begin{aligned} \lim \int _E|f_n-f|d\mu =0. \end{aligned}$$

(11.1c)

This gives an alternate proof of the Lebesgue dominated convergence theorem when $\mu (E)<\infty$. Give an example showing that the conclusion is false if E is not of finite measure.

11.4. :

Let $\{f_n\}$ be a sequence of integrable functions in E, satisfying (11.1c) for an integrable f. Then $\{f_n\}$ is uniformly integrable .

11.5. :

Show that the following sequences of functions defined in $(-1,1)$, are not uniformly Lebesgue integrable.

$$\begin{aligned} \begin{aligned} f_n(x)&=\sqrt{n}e^{-x^2}{n};\qquad&f_n(x)&= \frac{x}{\sqrt{n}}e^{-x^2}{n};\\ f_n(x)&=\frac{n}{n^2 x^2+1};\qquad&f_n(x)&=\frac{n^2 x}{(n^2 x^2+1)^2}. \end{aligned} \end{aligned}$$

(11.2c)

11.6. :

Let E be of finite measure and let $\{f_t\}:E\rightarrow \mathbb {R}^*$ be a family of functions uniformly integrable in E and such that $f_t(x)$ is continuous in $t\in (0,1)$ for every fixed $x\in E$. Prove that

$$\begin{aligned} (0,1)\ni t\rightarrow \int _Ef_td\mu \quad \text {is continuous}. \end{aligned}$$

 

12c Product of Measures

1.1 12.1c Product of a Finite Sequence of Measure Spaces

Given a finite sequence of measure spaces $\{X_j,\mathcal {A}_j,\mu _j\}_{j=1}^n$ for some $n\in \mathbb {N}$ their product space is constructed by the following steps:  

12.1. :

Measurable n -Rectangles. A measurable n -rectangle is a set of the form

$$\begin{aligned} E={\mathop {\textstyle {\prod }}\limits }_{j=1}^nE_j\quad \text { for }\quad E_j\in \mathcal {A}_j\quad \text { for all }\> j=1,\dots ,n. \end{aligned}$$

Denote by $\mathcal {R}_o^n$ be the collection of all measurable n-rectangles and prove that it forms a semi-algebra.

12.2. :

Measuring Measurable Rectangles. On $\mathcal {R}_o^n$ introduce the set function

$$\begin{aligned} \mathcal {R}_o^n\ni E\rightarrow \lambda _n(E)=\mathop {\textstyle {\prod }}\limits _{j=1}^n\mu _j(E_j). \end{aligned}$$

(12.1c)

Prove that $\lambda _n$ is countably additive on $\mathcal {R}_o^n$. For this use a n-dimensional version of Proposition 12.1. As a consequence $\lambda _n$ is well defined on $\mathcal {R}_o^n$ in the sense that, for $Q\in \mathcal {R}_o^n$, the value $\lambda _n(Q)$ is independent of a particular partition of elements in $\mathcal {R}_o^n$ making up Q.

12.3. :

Constructing the Product Measure Space. The set function $\lambda _n$ on $\mathcal {R}_o^n$ generates an outer measure $\mu _e^n$ on ${\mathop {\textstyle {\prod }}\limits }_{j=1}^n X_j$. The latter in turn generates the measure space $\{Y_n,\mathcal {B}_n,\nu _n\}$, where

$$\begin{aligned} Y_n=\mathop {\textstyle {\prod }}\limits _{j=1}^n X_j,\quad \mathcal {B}_n=\Big (\mathop {\textstyle {\prod }}\limits _{j=1}^n\mathcal {A}_j\Big ), \quad \nu _n=\Big (\mathop {\textstyle {\prod }}\limits _{j=1}^n\mu _j\Big ). \end{aligned}$$

Moreover $\mathcal {R}_o^n\subset \mathcal {B}_n$ and $\nu _n(E)=\lambda _n(E)$ for all $E\in \mathcal {R}_o^n$.

 

1.1.1 12.1.1c Alternate Constructions

Fix an integer $1\le m<n$ and construct first, by the indicated procedure, the two measure spaces $\{Y_m,\mathcal {B}_m,\nu _m\}$ and $\{Y^m,\mathcal {B}^m,\nu ^m\}$, where

$$\begin{aligned} \begin{array}{ccc} {\displaystyle Y_m=\mathop {\textstyle {\prod }}\limits _{j=1}^m X_j},\quad &{}{\displaystyle \mathcal {B}_m=\Big (\mathop {\textstyle {\prod }}\limits _{j=1}^m\mathcal {A}_j\Big )},\quad &{}{\displaystyle \nu _m=\Big (\mathop {\textstyle {\prod }}\limits _{j=1}^m\mu _j\Big )}\\ {}\\ {\displaystyle Y^m=\mathop {\textstyle {\prod }}\limits _{j=m+1}^n X_j}, \quad &{}{\displaystyle \mathcal {B}^m=\Big (\mathop {\textstyle {\prod }}\limits _{j=m+1}^n\mathcal {A}_j\Big )},\quad &{}{\displaystyle \nu ^m=\Big (\mathop {\textstyle {\prod }}\limits _{j=m+1}^n\mu _j\Big )}. \end{array} \end{aligned}$$

Then construct their product

$$\begin{aligned} \{Y_m\times Y^m, (\mathcal {B}_m\times \mathcal {B}^m), (\nu _m\times \nu ^m)\}. \end{aligned}$$

(12.2c)

Denote by $\mathcal {A}_o^n$ the smallest $\sigma$-algebra containing $\mathcal {R}_o^n$.  

12.4. :

Prove that $\mathcal {A}_o^n$ is contained in $(\mathcal {B}_m\times \mathcal {B}^m)$ for all $1\le m<n$.

12.5. :

Prove that for all $1\le m<n$, the restriction of $(\nu _m\times \nu ^m)$ to $\mathcal {R}_o^n$ coincides with the function $\lambda _n$ introduced in (12.1c).

12.6. :

Prove that for all $1\le m<n$, the product space in (12.2c) is generated by the same outer measure $\mu _e^n$, generated by $\lambda _n$ on $\mathcal {R}_o^n$. Conclude that

$$\begin{aligned} \{Y_n,\mathcal {B}_n,\nu _n\}=\{Y_m\times Y^m,(\mathcal {B}_m\times \mathcal {B}^m), (\nu _m\times \nu ^m)\} \end{aligned}$$

(12.3c)

and thus the finite product of measure spaces is associative.

12.7. :

Prove that the Lebesgue measure in $\mathbb {R}^N$ coincides with the product of N copies of the Lebesgue measure in $\mathbb {R}$.

 

13c On the Structure of $(\mathcal {A}\times \mathcal {B})$

 

13.1. :

Let $\{X,\mathcal {A},\mu \}$ and $\{Y,\mathcal {B},\nu \}$ be complete measure spaces. Let $A\subset X$ be $\mu$-measurable and let $B\subset Y$ be not $\nu$-measurable. Denote by $(\mathcal {A}\times \mathcal {B}) _e$ the outer measure on $\mathcal {A}\times \mathcal {B}$ that generates the product measure space $\{X\times Y,(\mathcal {A}\times \mathcal {B}) ,(\mu \times \nu )\}$. Prove that:  

i. :

If $(\mathcal {A}\times \mathcal {B}) _e(A\times B)=0$ then $A\times B$ is $(\mathcal {A}\times \mathcal {B})$-measurable.

ii. :

If $0<(\mathcal {A}\times \mathcal {B}) _e(A\times B)<\infty$ then $A\times B$ is not $(\mathcal {A}\times \mathcal {B})$-measurable. If $(\mathcal {A}\times \mathcal {B}) _e(A\times B)=\infty$ then $A\times B$ might be $(\mu \times \nu )$-measurable. Give an example or an argument.

iii. :

If $\{X,\mathcal {A},\mu \}$ and $\{Y,\mathcal {B},\nu \}$ are $\sigma$-finite and $(\mathcal {A}\times \mathcal {B}) _e(A\times B)>0$, then then $A\times B$ is not $(\mathcal {A}\times \mathcal {B})$-measurable.

 

13.2. :

Let $\{X,\mathcal {A},\mu \}$ be non complete and let $A\subset X$ be non $\mu$-measurable, but included in a $\mu$-measurable set $A^\prime$ of measure zero. For every $\nu$-measurable set $B\subset Y$, the rectangle $A\times B$ is $(\mu \times \nu )$-measurable. As a consequence, the assumption that both $\{X,\mathcal {A},\mu \}$ and $\{Y,\mathcal {B},\nu \}$ be complete, cannot be removed from Proposition 13.4.

13.3. :

Let $E\subset [0,1]$ be the Vitali non measurable set. The diagonal set $\mathcal {E}=\{(x,x)|x\in E\}$ is Lebesgue measurable in $\mathbb {R}^2$ and has measure zero. The rectangle $E\times E$ is not Lebesgue measurable in $\mathbb {R}^2$.

13.4. :

(Sierpinski [146]). Let $\mathbb {R}$ be well ordered by $\prec$, denote by $\varOmega$ the first uncountable and let

$$\begin{aligned} X=Y=E_\varOmega =\{x\in \mathbb {R}|x\prec \varOmega \}. \end{aligned}$$

Let $\mathcal {A}$ be the $\sigma$-algebra of the subsets of X that are either countable or their complement is countable. For $E\in \mathcal {A}$, let $\mu (E)=0$ if E is countable and $\mu (E)=1$ otherwise. Consider the set

$$\begin{aligned} E=\{(x,y)\in X\times X\bigm | x\prec y\}. \end{aligned}$$

All the x and y sections of E are measurable. However E is not $(\mu \times \mu )$-measurable. Indeed if E were measurable, it would have finite measure, since

$$\begin{aligned} E\subset X\times X \quad \text { and }\quad (\mu \times \mu )(X\times X))=\mu (X)\mu (X)=1. \end{aligned}$$

Therefore, it would have to satisfy Proposition 13.4. However

$$\begin{aligned} \int _Y\mu (E_y)d\nu =0\qquad \text { and }\qquad \int _X\mu (E_x)d\mu =1. \end{aligned}$$

 

1.1 13.1c Sections and Their Measure

Given a finite sequence $\{X_j,\mathcal {A}_,\mu _j\}_{j=1}^n$ of measure spaces, let $\{Y_n,\mathcal {B}_n,\nu _n\}$ be their product measure space as constructed either in (12.1c)–(12.3c).

For a set $E\subset Y_n$ and $y_m\in Y_m$, and $y^m\in Y^m$ for some $1\le m<n$, the $y_m$-section $E_{y_m}$ of E and the $y^m$-section $E_{y^m}$ of E are defined as

$$\begin{aligned} \begin{aligned} Y^m\ni E_{y_m}&=\{y^m\bigm |(y_m,y^m)\in E\> \text { for a fixed }\> y_m\in Y_m\}\\ Y_m\ni E_{y^m}&=\{y_m\bigm |(y_m,y^m)\in E\> \text { for a fixed }\> y^m\in Y^m\}. \end{aligned} \end{aligned}$$

(13.4c)

 

13.5. :

Prove that for all $E\in \mathcal {A}_o^n$ and all $1\le m<n$ the sections $E_{y_m}\in \mathcal {B}^m$ and $E_{y^m}\in \mathcal {B}_m$.

13.6. :

Prove that for all $E\in \mathcal {A}_o^n$ of finite $\nu _n$ measure and for all $1\le m<n$

$$\begin{aligned} \nu _n(E)=\int _{Y_n}\chi _E d\nu _n =\int _{Y_m}\nu ^m(E_{y_m}) d\nu _m =\int _{Y^m}\nu _m(E_{y^m}) d\nu ^m. \end{aligned}$$

(13.5c)

13.7. :

State and prove a version of Fubini’s Theorem when all the measure spaces $\{X_j,\mathcal {A}_j,\mu _j\}$, for $j=1,\dots ,n$, are complete.

13.8. :

State and prove a version of Tonelli’s Theorem when all the measure spaces $\{X_j,\mathcal {A}_j,\mu _j\}$, for $j=1,\dots ,n$, are complete and $\sigma$-finite.

 

14c The Theorem of Fubini–Tonelli

 

14.1. :

Let $\omega _N$ be the measure of the unit sphere in $\mathbb {R}^N$. Then

$$\begin{aligned} \omega _{N+1}=2\omega _N\int _0^{\pi /2} (\sin t)^{N-1}dt. \end{aligned}$$

14.2. :

By the Fubini–Tonelli theorem

$$\begin{aligned} \int _{\mathbb {R}^N}e^{-|x|^2}dx=\mathop {\textstyle {\prod }}\limits _{i=1}^N \int _{\mathbb {R}} e^{-x_i^2}dx_i=\pi ^{N/2}. \end{aligned}$$

14.3. :

Let $\{X,\mathcal {A},\mu \}$ be [0, 1] with the Lebesgue measure and let $\{Y,\mathcal {B},\nu \}$ be the rationals in [0, 1] with the counting measure. The function $f(x,y)=x$ is integrable on $\{X,\mathcal {A},\mu \}$ and not integrable on the product space.

14.4. :

Let [0, 1] be equipped with the Lebesgue measure $\mu$ and the counting measure $\nu$ both acting on the same $\sigma$-algebra of the Lebesgue measurable subsets of [0, 1]. The corresponding product measure space is not $\sigma$-finite since $\nu$ is not $\sigma$-finite. The diagonal set $E=\{x=y\}$ is of the type of $\mathcal {R}_{\sigma \delta }$, and hence $(\mu \times \nu )$-measurable and

$$\begin{aligned} \nu (E_x)=1\quad \forall \,x\in [0,1]\quad \text { and }\quad \mu (E_y)=0\quad \forall \,y\in [0,1]. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{[0,1]}\nu (E_x)d\mu =1 \quad \text { and }\quad \int _{[0,1]}\mu (E_y)d\nu =0. \end{aligned}$$

Moreover

$$\begin{aligned} \iint _{[0,1]\times [0,1]}\chi _{E}d(\mu \times \nu )=\infty . \end{aligned}$$

14.5. :

Let f and g be integrable functions on complete measure spaces $\{X,\mathcal {A},\mu \}$ and $\{Y,\mathcal {B},\nu \}$, respectively. Then $F(x,y)=f(x)g(y)$ is integrable on the product space $(X\times Y)$.

14.6. :

Let $\{X,\mathcal {A},\mu \}$ be any measure space and let $\{Y,\mathcal {B},\nu \}$ be $\mathbb {N}$ with the counting measure. Prove that the conclusion of Fubini’s theorem continues to hold even i $\{X,\mathcal {A},\mu \}$ is not complete, and that the conclusion of Tonelli’s theorem continues to hold even if $\{X,\mathcal {A},\mu \}$ is neither complete nor $\sigma$-finite.

14.7. :

Let $\{X,\mathcal {A},\mu \}$ and $\{Y,\mathcal {B},\nu \}$ be both $\mathbb {N}$ with the counting measure and consider the function

$$\begin{aligned} f(m,n)=\left\{ \begin{array}{rl} 1 &{} \quad \text { if }\> m=n\\ -1 &{} \quad \text { if }\> m=n+1\\ 0 &{} \quad \text { otherwise}. \end{array}\right. \end{aligned}$$

Prove that f is not integrable in the product space $\{X\times Y,(\mathcal {A}\times \mathcal {B}) ,(\mu \times \nu )\}$ and the conclusion of Fubini’s theorem fails.

14.8. :

Let $\{X,\mathcal {A},\mu \}$ and $\{Y,\mathcal {B},\nu \}$ be both (0, 1) with the Lebesgue measure. Let $f:(0,1)\rightarrow \mathbb {R}$ be measurable and assume that

$$\begin{aligned} (0,1)\times (0,1)\ni (x,y)\rightarrow f(x)-f(y) \end{aligned}$$

is integrable in the product space $\{X\times Y,(\mathcal {A}\times \mathcal {B}) ,(\mu \times \nu )\}$. Prove that f is integrable in (0, 1).

 

15c Some Applications of the Fubini–Tonelli Theorem

1.1 15.1c Integral of a Function as the “Area Under the Graph”

Let $\{X,\mathcal {A},\mu \}$ be complete, and $\sigma$-finite, and let $f:X\rightarrow \mathbb {R}^*$ be a nonnegative and integrable. The graph of f on E is the set of points $\{x:(x,f(x)\}$ and the set of points “under the graph” is

$$\begin{aligned} U_f=\{(x,y)\bigm | x\in X:\> 0\le y<f(x)\}. \end{aligned}$$

Prove that $U_f$ is measurable in the product measure space of $\{X,\mathcal {A},\mu \}$ and $\mathbb {R}$ with the Lebesgue measure dx, and

$$\begin{aligned} \{\text {measure of}\, U_f\}=\int _X f(x)d\mu . \end{aligned}$$

Prove that the graph of f as a subset of $X\times \mathbb {R}$ is measurable in the product measure and has measure zero.

1.2 15.2c Distribution Functions

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $f:X\rightarrow \mathbb {R}^*$ and $\{f_n\}:X\rightarrow \mathbb {R}^*$ be nonnegative and integrable. Prove:

Proposition 15.1c

Assume that $\mu ([f>t])\not \equiv \infty$. Then

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0}\mu ([f>t+\varepsilon ])= \mu ([f>t])\\&\lim _{\varepsilon \rightarrow 0}\mu ([f>t-\varepsilon ])= \mu ([f\ge t]). \end{aligned}$$

Therefore, the distribution function $t\rightarrow \mu ([f>t])$ is right-continuous, and it is continuous at a point t if only if $\mu ([f=t])=0$.

Proposition 15.2c

Let $\{f_n\}\rightarrow f$ in measure. Then for every $\varepsilon >0$

$$\begin{aligned}&\limsup \mu ([f_n>t]) \le \mu ([f>t-\varepsilon ])\\&\liminf \mu ([f_n>t]) \ge \mu ([f>t+\varepsilon ]). \end{aligned}$$

Therefore, $\mu ([f_n>t])\rightarrow \mu ([f>t])$ at those t where the distribution function of f is continuous.

 

15.1. :

For a measurable function f on $\{X,\mathcal {A},\mu \}$ set

$$\begin{aligned} \mathbb {R}^*\ni t\rightarrow f_*(t)=\mu \big ([f\le t]\big ) \end{aligned}$$

(15.1c)

prove that $f_*$ is nondecreasing, right-continuous, and $f_*(\infty )=\mu (X)$.

15.2. :

Give an example of f and $\{X,\mathcal {A},\mu \}$ for which $f_X$ is right-continuous and not continuous.

15.3. :

The function $f_*$ generates the Lebesgue–Stiltjies measure $\mu _{f_*}$ on $\mathbb {R}$. Let $\mu _f$ be the measure defined by (9.1c). Prove that $\mu _{f_*}=\mu _f$ on the Borel sets of $\mathbb {R}$ and that $\mu _{f_*}$ is the completion of $\mu _f$.

 

17c The Radon-Nikodým Theorem

 

17.1. :

Let $\nu$ be the Lebesgue measure in [0, 1] and let $\mu$ be the counting measure on the same $\sigma$-algebra of the Lebesgue measurable subsets of [0, 1]. Then $\nu$ is absolutely continuous with respect to $\mu$, but it does not exist a nonnegative, $\mu$-measurable function $f:[0,1]\rightarrow \mathbb {R}^*$ for which $\nu$ can be represented as in (17.1).

17.2. :

In [0, 1] let $\mathcal {A}$ be the $\sigma$-algebra of the sets that are, either countable or have countable complement. Let $\mu$ be the counting measure on $\mathcal {A}$ and let $\nu :\mathcal {A}\rightarrow \mathbb {R}^*$ be defined by $\nu (E)=0$ if E is countable and $\nu (E)=1$ otherwise. Then $\nu$ is absolutely continuous with respect to $\mu$ but it does not have a Radon-Nikodým derivative .

17.3. :

Changing the Variables of Integration: Let the assumptions of the Radon-Nikodým theorem hold. If g is $\nu$-integrable, $g(d\nu /d\mu )$ is $\mu$-integrable and for every measurable set E

$$\begin{aligned} \int _Eg d\nu =\int _Eg \frac{d\nu }{d\mu }d\mu . \end{aligned}$$

(17.1c)

17.4. :

Linearity of the Radon-Nikodým Derivative: Let $\mu$ and $\nu _1$ and $\nu _2$ be $\sigma$-finite measures defined on the same $\sigma$-algebra $\mathcal {A}$. If $\nu _1$ and $\nu _2$ are absolutely continuous with respect to $\mu$

$$\begin{aligned} \frac{d(\nu _1+\nu _2)}{d\mu }=\frac{d\nu _1}{d\mu }+ \frac{d\nu _2}{d\mu }\qquad \mu \text {-a.e.}. \end{aligned}$$

(17.2c)

17.5. :

The Chain Rule: Let $\mu$, $\nu$, and $\eta$ be $\sigma$-finite measures on X defined on the same $\sigma$-algebra $\mathcal {A}$. Assume that $\mu$ is absolutely continuous with respect to $\eta$ and that $\nu$ is absolutely continuous with respect to $\mu$. Then

$$\begin{aligned} \frac{d\nu }{d\eta }=\frac{d\nu }{d\mu }\frac{d\mu }{d\eta } \quad \text { a.e. with respect to}\, \eta . \end{aligned}$$

(17.3c)

17.6. :

Derivative of the Inverse: Let $\mu$ and $\nu$ be two not identically zero, $\sigma$-finite measures defined on the same $\sigma$-algebra $\mathcal {A}$ and mutually absolutely continuous. Then

$$\begin{aligned} \frac{d\nu }{d\mu }\ne 0\quad \text { and }\quad \frac{d\mu }{d\nu }=\left( \frac{d\nu }{d\mu }\right) ^{-1}. \end{aligned}$$

(17.4c)

Hint: For all nonnegative $\nu$-measurable functions g

$$\begin{aligned} \int _Eg d\nu =\int _Eg\frac{d\nu }{d\mu }d\mu . \end{aligned}$$

This is true for simple functions and hence for nonnegative $\nu$-measurable functions g. Now for all $E\in \mathcal {A}$ of finite $\mu$-measure

$$\begin{aligned} \mu (E)=\int _E1 d\mu =\int _E\frac{d\mu }{d\nu }d\nu =\int _E\frac{d\mu }{d\nu }\frac{d\nu }{d\mu }d\mu . \end{aligned}$$

The inverse formula (17.4c) follows from the uniqueness of the Radon-Nikodým derivative.

17.7. :

Let $\{X,\mathcal {A},\mu \}$ be [0, 1] with the Lebesgue measure. Let $\{E_n\}$ be a measurable partition of [0, 1] and let $\{\alpha _n\}$ be a sequence of positive numbers such that $\mathop {\textstyle {\sum }}\limits \alpha _n<\infty$. Find the Radon-Nikodým derivative of the measure

$$\begin{aligned} \mathcal {A}\ni E\longrightarrow \nu (E)=\mathop {\textstyle {\sum }}\limits \,\alpha _n\mu (E\cap E_n) \end{aligned}$$

with respect to the Lebesgue measure.

 

Proposition 17.1c

Let $\mu$ and $\nu$ be two measures defined on the same $\sigma$-algebra $\mathcal {A}$. Assume that $\nu$ is finite. Then $\nu$ is absolutely continuous with respect to $\mu$ if and only if for every $\varepsilon >0$ there exists $\delta >0$ such that $\nu (E)<\varepsilon$ for every set $E\in \mathcal {A}$ such that $\mu (E)<\delta$.

Proof

(Necessity) If not, there exist $\varepsilon >0$ and a sequence of measurable sets $\{E_n\}$, such that $\nu (E_n)\ge \varepsilon$ and $\mu (E_n)\le 2^{-n}$. Let $E=\limsup E_n$ and compute

$$\begin{aligned} \nu (E)=\nu (\limsup E_n)\ge \limsup \nu (E_n)\ge \varepsilon . \end{aligned}$$

On the other hand for all $n\in \mathbb {N}$

$$\begin{aligned} \mu (E)\le \mu \big (\mathop {\textstyle {\bigcup }}\limits _{j=n}^\infty \,E_j\big )\le \mathop {\textstyle {\sum }}\limits _{j=n}^\infty \,\mu (E_j)\le \frac{1}{2^{n-1}}. \end{aligned}$$

$\blacksquare$

 

17.8. :

The proposition might fail if $\nu$ is not finite. Let $X=\mathbb {N}$ and for every subset $E\subset \mathbb {N}$, set

$$\begin{aligned} \mu (E)=\mathop {\textstyle {\sum }}\limits _{n\in E}\,\frac{1}{2^n}\qquad \nu (E)=\mathop {\textstyle {\sum }}\limits _{n\in E}\,2^n. \end{aligned}$$

17.9. :

More on $\varvec{\nu \ll \mu }$ : There exist $\sigma$-finite measures $\nu$ on the Lebesgue measurable sets of $\mathbb {R}$, absolutely continuous with respect to the Lebesgue measure $\mu$ on $\mathbb {R}$ and such that $\nu (E)=\infty$ for every measurable set E with nonempty interior. Let $\{r_n\}$ be the sequence of the rationals in $\mathbb {R}$ and set

$$\begin{aligned} g(x)=\mathop {\textstyle {\sum }}\limits \frac{1}{2^{n}}\frac{e^{-|x|}}{|x-r_n|}\quad \text { and } \quad \nu (E)=\int _Egd\mu \end{aligned}$$

for all Lebesgue measurable set $E\subset \mathbb {R}$.

17.10. :

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $\nu$ be a signed measure on $\mathcal {A}$ of finite total variation $|\nu |$. Then $\nu \ll \mu$ if and only if for every $\varepsilon >0$ there exists $\delta >0$ such that $|\nu |(E)<\varepsilon$ for all $E\in \mathcal {A}$ such that $\mu (E)<\delta$. Give an counterexample if $|\nu |$ is not finite.

 

18c A Proof of the Radon-Nikodým Theorem When Both $\mu$ and $\nu$ are $\sigma$-Finite

A more constructive proof of Theorem 17.1 can be given if both $\mu$ and $\nu$ are $\sigma$-finite. Assume first that both $\mu$ and $\nu$ are finite. Let $\varPhi$ be the family of all measurable nonnegative functions $\varphi :X\rightarrow \mathbb {R}^*$ such that

$$\begin{aligned} \int _E\varphi d\mu \le \nu (E)\quad \text { for all } \> E\in \mathcal {A}. \end{aligned}$$

Since $0\in \varPhi$ such a class is not empty. For two given functions $\varphi _1$ and $\varphi _2$ in $\varPhi$ the function $\max \{\varphi _1;\varphi _2\}$ is in $\varPhi$. Indeed for any $E\in \mathcal {A}$

$$\begin{aligned} \int _E\max \{\varphi _1;\varphi _2\}d\mu&= \int _{E\cap [\varphi _1\ge \varphi _2]}\varphi _1d\mu + \int _{E\cap [\varphi _2>\varphi _1]}\varphi _2d\mu \\&\le \nu (E\cap [\varphi _1\ge \varphi _2])+ \nu (E\cap [\varphi _2<\varphi _1])=\nu (E). \end{aligned}$$

Since $\nu$ is finite

$$\begin{aligned} M=\sup _{\varphi \in \varPhi }\int _X\varphi d\mu \le \nu (X)<\infty . \end{aligned}$$

Let $\{\varphi _n\}$ be a sequence of functions in $\varPhi$ such that

$$\begin{aligned} \lim \int _X \varphi _nd\mu =M \end{aligned}$$

and set $f_n=\max \{\varphi _1,\cdots ,\varphi _n\}$. The sequence $\{f_n\}$ is nondecreasing, and $\mu$-a.e. convergent to a function f that belongs to $\varPhi$. Indeed by monotone convergence, for every measurable set E

$$\begin{aligned} \int _Efd\mu =\lim \int _Ef_nd\mu \le \nu (E). \end{aligned}$$

Such a limiting function is the f claimed by the theorem. For this it suffices to establish that the measure

$$\begin{aligned} \mathcal {A}\ni E\longrightarrow \eta (E) =\nu (E)-\int _Efd\mu \end{aligned}$$

is identically zero. If not, there exists a set $A\in \mathcal {A}$ such that $\eta (A)>0$. Since both $\nu$ and $\eta$ are absolutely continuous with respect to $\mu$, for such a set, $\mu (A)>0$. Also, since $\mu$ is finite, there exists $\varepsilon >0$ such that

$$\begin{aligned} \xi (A)=\eta (A)-\varepsilon \mu (A)>0. \end{aligned}$$

The set function

$$\begin{aligned} \mathcal {A}\ni E\longrightarrow \xi (E)=\eta (E)-\varepsilon \mu (E) \end{aligned}$$

is a signed measure on $\mathcal {A}$. Therefore, by Proposition 16.1 the set A contains a positive subset $A_o$ of positive measure. In particular

$$\begin{aligned} \eta (E\cap A_o)-\varepsilon \mu (E\cap A_o) \ge 0\quad \text { for all }\> E\in \mathcal {A}. \end{aligned}$$

From this and the definition of $\eta$

$$\begin{aligned} \varepsilon \mu (E\cap A_o)\le \nu (E\cap A_o)- \int _{E\cap A_o}fd\mu \quad \text { for all }\> E\in \mathcal {A}. \end{aligned}$$

The function $(f+\varepsilon \chi _{A_o})$ belongs to $\varPhi$. Indeed, for every measurable set E

$$\begin{aligned} \int _E(f+\varepsilon \chi _{A_o})d\mu&=\int _{E-A_o}fd\mu + \int _{E\cap A_o}(f+\varepsilon )d\mu \\&\le \nu (E-A_o)+\nu (E\cap A_o)=\nu (E). \end{aligned}$$

This however contradicts the definition of M since

$$\begin{aligned} \int _X(f+\varepsilon \chi _{A_o})d\mu =M+\varepsilon \mu (A_o)>M. \end{aligned}$$

If $g:X\rightarrow \mathbb {R}^*$ is another nonnegative measurable function by which the measure $\nu$ can be represented, let $A_n=[f-g\ge \frac{1}{n}]$. Then for all $n\in \mathbb {N}$

$$\begin{aligned} \frac{1}{n}\mu (A_n)\le \int _{A_n}(f-g)d\mu = \nu (A_n)-\nu (A_n)=0. \end{aligned}$$

Thus $f=g$ $\mu$-a.e. in X.

Assume next that $\mu$ is $\sigma$-finite and $\nu$ is finite. Let $E_n$ be a sequence of measurable, expanding sets such that

$$\begin{aligned} \mu (E_n)\le \mu (E_{n+1})<\infty \quad \text { and }\quad X=\mathop {\textstyle {\bigcup }}\limits E_n \end{aligned}$$

and denote by $\mu _n$ the restriction of $\mu$ to $E_n$. Let $f_n$ be the unique function claimed by the Radon-Nikodým theorem for the pair of finite measures $\{\mu _n;\nu \}$. By construction $f_{n+1}\bigm |_{E_n}= f_n$ for all n. The function f claimed by the theorem is $f=\sup f_n$. Indeed if $E\in \mathcal {A}$, by monotone convergence

$$\begin{aligned} \nu (E)=\lim \nu (E\cap E_n)= \lim \int _Ef_nd\mu =\int _Efd\mu . \end{aligned}$$

The uniqueness of such a f is proved as in the case of $\mu$ finite. Finally a similar argument, establishes the theorem when also $\nu$ is $\sigma$-finite.$\blacksquare$