7 Integration

Definition

Let $(X,\mathcal{A},\mu )$ be a measure space.

• Let $s={\sum }_{i=1}^n{a}_i{\chi }_{E_i}$ be a nonnegative measurable simple function. The (Lebesgue) integral of $s$ over $X$ is the number

$${\int }_{X}sd\mu =\sum _{i=1}^n{a}_i\mu \left(E_i\right)$$

• If $f\ge 0$ is a measurable function. The integral of f over $X$ is the number ${\int }_{X}fdu$ defined by

$${\int }_{X}fd\mu :=\underset{s\le f}{\sup }{\int }_{X}sd\mu$$

where the supremum is taken over all measurable simple functions $s:X\to \mathbb{R}$ satisfying $0\le s(x)\le f(x)$ for all $x\in X$.

Let $f$ be measurable and let $f^{+}=\max \{f,0\}$ and $f^{-}=\max \{-f,0\}$. Provided ${\int }_X{f}^{+}d\mu$ and ${\int }_X{f}^{-}d\mu$ are not both infinite, define

$${\int }_{X}fd\mu ={\int }_X{f}^{+}d\mu -{\int }_X{f}^{-}d\mu$$

Finally, if $f=u+iv$ is complex-valued and ${\int }_X(|u|+|v|)d\mu$ is finite, define

$${\int }_{X}fd\mu ={\int }_{X}ud\mu +i{\int }_{X}vd\mu .$$

Definition

If $f$ is measurable and ${\int }_X|f|d\mu <\infty$, we say $f$ is integrable.

Proposition

There are some easy propositions, see here.

MCT

monotone convergence theorem

Suppose $\left\{f_n\right\}$ is a sequence of non-negative measurable functions with $f_1(x)\le f_2(x)\le \cdots$ for all $x$ and with ${\lim }_{n\to \infty }{f}_n(x)=f(x)$ for all $x$. Then ${\int }_X{f}_{n}d\mu \to {\int }_{X}fd\mu$.

\begin{proof} Let $s$ be a simple function with $0⩽s⩽f$. For any fixed $c\in (0,1)$, let $A_n=\{x:f_n(x)⩾cs(x)\}$. We have $A_1\subseteq A_2\subseteq \cdots$ and ${\cup }_{n=1}^{\infty }{A}_n=X$. Hence, ${\lim }_{n\to \infty }\mu (A_n)=\mu (X)$. Note that

$${\int }_X{f}_n⩾{\int }_{A_n}{f}_n⩾{\int }_{A_n}csd\mu =c\sum a_j\mu (A_n\cap E_j)$$

where $s={\sum }_{j=1}^{\infty }{a}_j{\chi }_{E_j}$. As $n\to \infty$,

$$c\sum a_j\mu (A_n\cap E_j)\to c\sum a_j\chi (E_j)=c{\int }_{X}sd\mu .$$

Take $c\to 1$, then we have ${\lim }_{n\to \infty }{\int }_X{f}_n⩾{\int }_{X}s$ for all simple functions $s⩽f$ hold. Then we obtain ${\lim }_{n\to \infty }{\int }_X{f}_n⩾{\int }_{X}f$. On the other hand, as $f_1\uparrow f$, ${\lim }_{n\to \infty }{\int }_X{f}_n⩽{\int }_{X}f$. Now we finish the proof. \end{proof}

Remark. $f$ is non-negative is necessary. Let $X=[0,\infty )$ and $f_n(x)=-1/n$ for all $x$. Then $\int f_n=-\infty$, but $f_n\uparrow f$ where $f=0$ and $\int f=0$.

Additivity

Proposition

If $s,t⩾0$ are simple functions, then ${\int }_X(s+t)d\mu ={\int }_{X}sd\mu +{\int }_{X}td\mu$.

\begin{proof} Assume that $s={\sum }_{j=1}^n{a}_j{\chi }_{E_j}$ and $t={\sum }_{i=1}^m{b}_i{\chi }_{E_i}$. Then

$${\int }_X(s+t)d\mu =\sum _{i=1}^m\sum _{j=1}^n(a_i+b_j){\chi }_{A_i\cap B_j}={\int }_{X}sd\mu +{\int }_{X}td\mu .$$

We have done. \end{proof}

Theorem

If $f$ and $g$ are non-negative and measurable, then

\int_X(f+g) d \mu=\int_X f d \mu+\int_X g d \mu.

`\begin{proof}` By [[6 Convergence of Measurable Functions#^n03bsd|^n03bsd]], there exist $0\leqslant s_1\leqslant s_2\leqslant\cdots$ and $0\leqslant t_1\leqslant t_2\leqslant\cdots$ such that $s_i(x)\to f(x)$ and $t_i(x)\to g(x)$. Then by [[#^33d8fd|^33d8fd]], $\int_X s_i+t_i=\int_X s_i+\int_X t_i$. It follows that

\int_X(f+g)=\lim_{n\to\infty}\int_X(s_n+t_n)=\lim_{n\to\infty}(\int_Xs_n+\int_X t_n)=\int_Xf+\int_Xg.

Therefore, we have that $\int_X(f+g) d \mu=\int_X f d \mu+\int_X g d \mu$. `\end{proof}` > [!theorem] > > If $f$ and $g$ are integrable, then > > $$ \int_X(f+g) d \mu=\int_X f d \mu+\int_X g d \mu.

\begin{proof} Since $\int |f+g|⩽\int |f|+\int |g|<\infty$, then $f+g$ is integrable. Write

$$(f+g{)}^{+}-(f+g{)}^{-}=f+g=f^{+}-f^{-}+g^{+}-g^{-}$$

so that

$$(f+g{)}^{+}+f^{-}+g^{-}=f^{+}+g^{+}+(f+g{)}^{-}$$

Using the result for non-negative functions,

$$\int (f+g{)}^{+}+\int f^{-}+\int g^{-}=\int f^{+}+\int g^{+}+\int (f+g{)}^{-}$$

Rearranging,

$$\begin{array}{rl}\int (f+g)& =\int (f+g{)}^{+}-\int (f+g{)}^{-}\\ & =\int f^{+}-\int f^{-}+\int g^{+}-\int g^{-}=\int f+\int g\end{array}$$

If $f$ and $g$ are complex-valued, apply the above to the real and imaginary parts. \end{proof}

Proposition

Let $s⩾0$ be a simple function. Assume that $\{E_i\}\subseteq \mathcal{A}$ with $E_i\cap E_j=\varnothing$ if $i\ne j$. Then ${\int }_{{\cup }_{i=1}^{\infty }{E}_i}sd\mu ={\sum }_{i=1}^{\infty }{\int }_{E_i}sd\mu$.

\begin{proof} Let $E={\cup }_{i=1}^{\infty }{E}_i$ and $F_n={\cup }_{i=1}^n{E}_i$. Then $F_n\uparrow E$ and

$${\int }_{E}s={\int }_{X}s{\chi }_E=\underset{n\to \infty }{\lim }{\int }_{X}s{\chi }_{F_n}=\underset{n\to \infty }{\lim }{\int }_{F_n}s.$$

Since ${\int }_{F_n}s={\sum }_{i=1}^n{\int }_{E_i}s$, we have that ${\int }_{{\cup }_{i=1}^{\infty }{E}_i}s={\sum }_{i=1}^{\infty }{\int }_{E_i}s$. \end{proof}

Proposition

If $f:X\to [0,\infty ]$ is measurable and $E_1,E_2,E_3,\dots$ is a sequence of pairwise disjoint measurable sets. Then

$${\int }_{{\cup }_{i=1}^{\infty }{E}_i}fd\mu =\sum _{i=1}^{\infty }{\int }_{E_i}fd\mu$$

\begin{proof} Let $E={\cup }_{i=1}^{\infty }{E}_i,f_n={\sum }_{k=1}^{n}f{\chi }_{E_k}$, then $f_n\uparrow f{\chi }_E$ and

$${\int }_X{f}_{n}d\mu =\sum _{k=1}^n{\int }_{E_k}fd\mu .$$

Taking $n\to \infty$ and using the monotone convergence theorem, one gets ${\int }_{{\cup }_{i=1}^{\infty }{E}_i}fd\mu ={\sum }_{i=1}^{\infty }{\int }_{E_i}fd\mu$. \end{proof}

Proposition

Let $f_n:X\to [0,\infty ]$ be a sequence of measurable functions. Then

$${\int }_E\sum _{n=1}^{\infty }{f}_n=\sum _{n=1}^{\infty }{\int }_E{f}_n$$

for all $E\in \mathcal{A}$.

\begin{proof} Let $F_n={\sum }_{i=1}^n{f}_i$. For any $a\in \mathbb{R}$, since $F_n\uparrow {\sum }_{n=1}^{\infty }{f}_n$, we have

$$\{x\in X:F(x)>a\}={\cup }_{n=1}^{\infty }\left\{x\in X:F_n(x)>a\right\}$$

Thus ${\sum }_{n=1}^{\infty }{f}_n$ is measurable and

$$\begin{array}{rl}{\int }_E\sum _{i=1}^{\infty }{f}_{i}d\mu & ={\int }_E\underset{n\to \infty }{\lim }\sum _{i=1}^n{f}_{i}d\mu ={\int }_E\underset{n\to \infty }{\lim }{F}_{n}d\mu \\ & =\underset{n\to \infty }{\lim }{\int }_E{F}_{n}d\mu =\sum _{i=1}^{\infty }{\int }_E{f}_{n}d\mu .\end{array}$$

Now we finish the proof. \end{proof}

Example. For a non-negative double sequence of reals, the order of summation can be reversed. See here.

Fatou Lemma and DCT

Fatou Lemma

Suppose the $f_n$ are non-negative and measurable. Then

$$\int \underset{n\to \infty }{\mathrm{liminf}}{f}_n⩽\underset{n\to \infty }{\mathrm{liminf}}\int f_n$$

\begin{proof} Note that

$${\int }_X\underset{n\to \infty }{\mathrm{liminf}}{f}_n={\int }_X\underset{n\to \infty }{\lim }\underset{k⩾n}{\inf }{f}_k=\underset{n\to \infty }{\lim }{\int }_X\underset{k⩾n}{\inf }{f}_k=\underset{n\to \infty }{\mathrm{liminf}}{\int }_X\underset{k⩾n}{\inf }{f}_k⩽\underset{n\to \infty }{\mathrm{liminf}}{\int }_X{f}_n.$$

We have done. \end{proof}

Remark. The inequality can be strict, see here.

Dominated convergence theorem

Suppose that $f_n,f$ are measurable extended real-valued functions and $f_n(x)\to f(x)$ for each $x$. Suppose there exists a non-negative integrable function $g$ such that $\left|f_n(x)\right|\le g(x)$ for all $x$. Then

$$\underset{n\to \infty }{\lim }\int f_{n}d\mu =\int fd\mu$$

\begin{proof} Since $f_n+g\ge 0$, by Fatou lemma,

$$\int f+\int g=\int (f+g)\le \underset{n\to \infty }{\mathrm{liminf}}\int \left(f_n+g\right)=\underset{n\to \infty }{\mathrm{liminf}}\int f_n+\int g$$

Since $g$ is integrable, $\int f\le {\mathrm{liminf}}_{n\to \infty }\int f_n(\ast )$. Similarly, $g-f_n\ge 0$, so

$$\int g-\int f=\int (g-f)\le \underset{n\to \infty }{\mathrm{liminf}}\int \left(g-f_n\right)=\int g+\underset{n\to \infty }{\mathrm{liminf}}\int \left(-f_n\right),$$

and so $-\int f\le {\mathrm{liminf}}_{n\to \infty }\int \left(-f_n\right)=-{\mathrm{limsup}}_{n\to \infty }\int f_n$, which with $(\ast )$ proves the theorem. \end{proof}

GDCT/An Extension of the Dominated Convergence

Given a measure space $(X,\mathcal{A},\mu )$. Let $\left\{f_n\right\}$ and $f$ be extended real-valued measurable functions and let $\left\{g_n\right\}$ and $g$ be nonnegative extended real-valued measurable functions on X. Suppose

• $f_n\to f$ and $g_n\to g$ a.e. on $X$.
• $g_n$ and $g$ are all integrable and ${\int }_X{g}_{n}d\mu \to {\int }_{X}gd\mu$.
• $\left|f_n\right|\le g_n$ on $X$ for any $n\in \mathbb{N}$.

Then $f$ is integrable and ${\int }_X{f}_{n}d\mu \to {\int }_{X}fd\mu$.

\begin{proof} Firstly, we show that $f$ is integrable. Note that

$$\begin{array}{rl}{\int }_X|f|d\mu & ={\int }_X\underset{n\to \infty }{\lim }\left|f_n\right|d\mu \\ & \le \underset{n\to \infty }{\mathrm{liminf}}{\int }_X\left|f_n\right|d\mu \\ & \le \underset{n\to \infty }{\mathrm{liminf}}{\int }_X{g}_{n}d\mu \\ & ={\int }_{X}gd\mu <\infty ,\end{array}$$

then $f$ is integrable.

Since $f_n+g_n⩾0$, then

$$\int (g+f)\le \underset{n\to \infty }{\mathrm{liminf}}\int \left(g_n+f_n\right)=\int g+\underset{n\to \infty }{\mathrm{liminf}}\int f_n$$

and so $\int f⩽\mathrm{liminf}\int f_n$. Similarly, since $g_n-f_n⩾0$ we can show that $\mathrm{limsup}\int f_n⩽\int f$. Now we have that $\lim \int f_n=\int f$. \end{proof}

Chebyshev’s Inequality

Chebyshev's Inequality

Let $(X,\mathcal{A},\mu )$ be a measure space $f$ and $\lambda$ a positive real number.

• If $f$ a nonnegative measurable function on $X$, then

$$\mu (\{x\in X,f(x)>\lambda \})\le \frac{1}{\lambda }{\int }_{X}fd\mu .$$

• If $g$ is an integrable function on $X$, then

$$\mu (\{x\in X,|g(x)|>\lambda \})\le \frac{1}{\lambda }{\int }_{\{x\in X,|g(x)|>\lambda \}}|g|d\mu .$$

\begin{proof} Easy. \end{proof}

Proposition

Let $(X,\mathcal{A},\mu )$ be a measure space and $f$ a nonnegative measurable function on $X$ for which ${\int }_{X}fd\mu <\infty$. Then $f$ is finite a.e. on $X$ and $\{x\in X:f(x)>0\}$ is $\sigma$-finite.

\begin{proof} By ^ceec71. \end{proof}

Proposition

Suppose $f$ is measurable and non-negative and ${\int }_{X}fd\mu =0$. Then $f=0$ almost everywhere.

\begin{proof} Since $\{x:f(x)>0\}={\cup }_{n=1}^{\infty }\{x:f(x)>1/n\}$, we have $0={\int }_{X}f⩾1/n\cdot \mu (\{x:f(x)>1/n\})$ and so $\mu (\{x:f(x)>0\})=0$. \end{proof}

Proposition

Suppose $f$ is real-valued and integrable and for every measurable set $A$ we have ${\int }_{A}fd\mu =0$. Then $f=0$ almost everywhere.

\begin{proof} Let $A=\{x:f(x)>0\}$, and let $B=\{x:f(x)<0\}$. By ^ra34bw, $f=0$ almost everywhere on both $A$ and $B$. Now we finish the proof. \end{proof}

VCT

Definition

Let $(X,\mathcal{A},\mu )$ be a measure space and $\left\{f_n\right\}$ a sequence of functions on $X$, each of which is integrable over $X$. The sequence $\left\{f_n\right\}$ is said to be uniformly integrable over $X$ provided for each $ϵ>0$, there is a $\delta >0$ such that for any natural number $n$ and measurable subset $E$ of $X$, if $\mu (E)<\delta$, then

$${\int }_E\left|f_n\right|d\mu <ϵ$$

The sequence $\left\{f_n\right\}$ is said to be tight over $X$ provided for each $ϵ>0$, there is a subset $X_0\subset X$ that has finite measure and, for any natural number $n$,

$${\int }_{X-X_0}\left|f_n\right|d\mu <ϵ$$

Lemma

Let $(X,\mathcal{A},\mu )$ be a measure space and the function $f$ be integrable over $X$. Then for each $ϵ>0$, there is a $\delta >0$ such that for any measurable subset $E$ of $X$,

$$\mu (E)<\delta \Rightarrow {\int }_E|f|d\mu <ϵ$$

Furthermore, for each $ϵ>0$, there is a subset $X_0\subset X$ that has finite measure and

$${\int }_{X-X_0}|f|d\mu <ϵ$$

\begin{proof} Assume that $f⩾0$ and ${\int }_{X}f<\infty$.

i) For any $ϵ>0$, let $s$ be a non-negative simple measurable function with $s⩽f$, and ${\int }_{X}f⩽{\int }_{X}s+ϵ/2$. Note that $s$ is a bounded function, then

$${\int }_{A}f⩽{\int }_A(f-s)+{\int }_{A}s⩽ϵ/2+M\mu (A).$$

Take $\delta =\mu (A)/2M$, then ${\int }_{A}f<ϵ$.

ii) Let $X_0=\{x:s(x)⩾0\}$, and let $s$ be the same function defined in i). Then $\mu (X_0)<\infty$, and

$${\int }_{X-X_0}f={\int }_{X-X_0}(f-s)+{\int }_{X-X_0}s⩽{\int }_X(f-s)<ϵ.$$

Now we finish the proof. \end{proof}

Vitali Convergence Theorem

Let $(X,\mathcal{A},\mu )$ be a measure space and $\left\{f_n\right\}$ a sequence of functions on $X$ that is both uniformly integrable and tight over $X$. Assume $f_n\to f$ pointwise a.e. on $X$ and the function $f$ is integrable over $X$. Then

\lim _{n \rightarrow \infty} \int_X f_n=\int_X f.

`\begin{proof}` For any $\epsilon>0$, take $X_0\in A$ with $\mu(X_0)<\infty$, then $\int_{X-X_0}(|f_n|+|f|)<\epsilon/6$ for all $n\in \mathbb{N}_+$. Further, there exists $\delta>0$ such that if $\mu(A)<\delta$, then $\int_A(|f_n|+|f|)<\epsilon/6$ for all $n\in \mathbb{N}_+$. Since $f_n\to f$ pointwise on $X_0$ and $\mu(X_0)<\epsilon$, by [[6 Convergence of Measurable Functions#^6ijlum|Egorov's theorem]], there exists $X_1\subseteq X_0$ with $\mu(X_0-X_1)<\delta$ such that $f_n\to f$ uniformly on $X_1$. It follows that

\begin{aligned} \int_X|f_n-f|&\leqslant\int_{X-X_0}(|f_n|+|f|)+\int {X_0}|f_n-f|\ &<\frac{\epsilon}{6}+\int{X_0-X_1}|f_n-f|+\int_{X_1}|f_n-f|\ &<\frac{\epsilon}{3}+\int_{X_1}|f_n-f|. \end{aligned}

As $f_n\to f$ uniformly on $X_1$, there exists $N$ such that $|f_n(x)-f(x)|<\epsilon/6\mu(X_1)$ for any $n>N$. Hence, when $n>N$, $\int_X|f_n-f|<\epsilon/2$, i.e., $\lim_{n\to\infty}\int_Xf_n=\int_X f$. `\end{proof}` > [!theorem] > > A bounded real-valued function $f$ on $[a, b]$ is Riemann integrable if and only if the set of points at which $f$ is discontinuous has Lebesgue measure 0, and in that case, $f$ is Lebesgue measurable and the Riemann integral of $f$ is equal in value to the Lebesgue integral of $f$.