Integration

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Fix a measurable space $(X,\mathcal{B})$. A measure on $(X,\mathcal{B})$ allows us to assign sizes to subsets of $X$. In this section we are going to understand how measures give rise to integrals. The key is the we can associate to each measurable set a function on $X$ whose integral can reasonably be thought of as the measure of $X$. We will then integrate more complicated functions by approximations.

Simple functions

Fix a set $X$ and $A\subset X$. The indicator function of $A$ is the function $$1_A(x)=\{\begin{array}{ll}1& x\in A\\ 0& x\notin A\end{array}$$ from $X$ to $\mathbb{R}$.

Lemma

Fix a measurable space $(X,\mathcal{B})$. For every $B\in \mathcal{B}$ the function $1_B:X\to \mathbb{R}$ is $(\mathcal{B},\mathcal{P}(\mathbb{R}))$ measurable.

Proof:

Fix $E\subset \mathbb{R}$. The set $(1_B{)}^{-1}(E)$ can only be $\varnothing$, $B$, $X\setminus B$ or $X$ according to which of $0,1$ belong to $E$. All four sets belong to $\mathcal{B}$ because $B\in \mathcal{B}$. Therefore $1_B$ is $(\mathcal{B},\mathcal{P}(\mathbb{R}))$ measurable.

Definition

Fix a measurable space $(X,\mathcal{B})$. A function $f:X\to \mathbb{R}$ is simple if $$f=a_1{1}_{B(1)}+\cdots +a_k{1}_{B(k)}$$ for sets $B(1),\dots ,B(k)$ in $\mathcal{B}$ and numbers $a_1,\dots ,a_k$.

The expression $$a_1{1}_{B(1)}+\cdots +a_k{1}_{B(k)}$$ is not unique. Indeed, when $B(1)$ and $B(2)$ are disjoint we can write $$1_{B(1)}+1_{B(2)}=1_{B(1)\cup B(2)}$$ and when $B(1)\cap B(2)$ is non-empty we can write $$a_1{1}_{B(1)}+a_2{1}_{B(2)}=a_1{1}_{B(1)\setminus B(2)}+(a_1+a_2)1_{B(1)\cap B(2)}+a_2{1}_{B(2)\setminus B(1)}$$ however, it is always possible to write a simple function in the form $$a_1{1}_{B(1)}+\cdots +a_k{1}_{B(k)}$$ where the sets $B(1),\dots ,B(k)$ are disjoint.

Lemma

A function $f:X\to \mathbb{R}$ is simple if and only if its range is finite.

Proof:

If $f$ can be written as $$a_1{1}_{B(1)}+a_2{1}_{B(2)}+\cdots +a_k{1}_{B(k)}$$ then its range is finite, as $f$ can only take values obtained from adding together at most $k$ of the terms $a_1,\dots ,a_k$. On the other hand, if the only values $f$ takes are $c_1,\dots ,c_k$ then $$f=c_1{1}_{f^{-1}(\{c_1\})}+\cdots +c_k{1}_{f^{-1}(\{c_k\})}$$ expresses $f$ as a simple function.

Integrating simple functions

Fix a measure space $(X,\mathcal{B},\mu )$. We define the integral of a simple, measurable function $f:X\to \mathbb{R}$ with respect to $\mu$ to be $$\int f\mathsf{d}\mu =\sum _{t\in \mathbb{R}}t\cdot \mu (f^{-1}(\{t\}))=\sum _{i=1}^n{a}_i\cdot \mu (f^{-1}(\{a_i\}))$$ where $a_1,\dots ,a_n$ are the values that $f$ takes.

We will first check that the integral of simple functions is linear.

Theorem

Let $(X,\mathcal{B},\mu )$ be a measure space. Let $f,g:X\to \mathbb{R}$ be simple and measurable. Then $$\begin{array}{c}\int f+g\mathsf{d}\mu =\int f\mathsf{d}\mu +\int g\mathsf{d}\mu \\ \alpha \int f\mathsf{d}\mu =\int \alpha f\mathsf{d}\mu \end{array}$$ for all $\alpha \in \mathbb{R}$.

Proof:

The second property is immediate. For the first, write $$f=\sum _{i=1}^m{a}_i{1}_{A(i)}g=\sum _{j=1}^r{b}_j{1}_{B(j)}$$ with $A(1),\dots ,A(m)$ pairwise disjoint and $B(1),\dots ,B(r)$ also pairwise disjoint. We can write $$A(i)=\bigcup _{j=1}^{r}A(i)\cap B(j)B(j)=\bigcup _{i=1}^{m}B(j)\cap A(i)$$ with both unions consisting of pairwise disjoint collections. Thus $$\begin{array}{c}f(x)=\sum _{i=1}^m\sum _{j=1}^r{a}_i{1}_{A(i)\cap B(j)}\\ g(x)=\sum _{j=1}^r\sum _{i=1}^m{b}_i{1}_{B(j)\cap A(i)}\\ (f+g)(x)=\sum _{i=1}^m\sum _{j=1}^r(a_i+b_j)1_{A(i)\cap B(j)}\end{array}$$ and adding the integrals of the first two simple functions gives the integral of the third simple function. ▮

Integrating non-negative functions

Now that we know how to integrate simple functions, we move on to non-negative measurable functions.

Definition

Fix a measure space $(X,\mathcal{B},\mu )$ and a measurable function $f:X\to [0,\mathrm{\infty }]$. Define the integral of $f$ with respect to $\mu$ to be $$\int f\mathsf{d}\mu =sup\{\int \varphi \mathsf{d}\mu :\varphi \text{ a simple function with }0\le \varphi \le f\}$$ the idea being to approximate the area under the graph of $f$ by the integral of a simple function.

Proposition

Fix $f,g:X\to [0,\mathrm{\infty }]$ measurable.

1. If $f\le g$ then ${\displaystyle \int f\mathsf{d}\mu \le \int g\mathsf{d}\mu }$
2. If $\alpha \ge 0$ then ${\displaystyle \alpha \int f\mathsf{d}\mu =\int \alpha f\mathsf{d}\mu }$

Proof:

The first of these is immediate because every simple function $0\le \varphi \le f$ satisfies $0\le \varphi \le g$. For the second, if $\alpha =0$ both sides are zero, and if $\alpha >0$ then the result follows because it is true for simple functions. ▮

We next want to prove that the integral is linear. We will not do this directly, instead proving first the fundamental convergence result on integration with respect to a measure.

Monotone convergence theorem

Theorem

For any measure space $(X,\mathcal{B},\mu )$ and any non-decreasing sequence $f_n:X\to [0,\mathrm{\infty }]$ of measurable functions the equality $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu =\int \underset{n\to \mathrm{\infty }}{lim}{f}_n\mathsf{d}\mu$$ holds.

Proof:

Define $g$ to be the pointwise limit of the sequence $n\mapsto f_n$ of functions. It is well-defined and measurable as a function from $X$ to $[0,\mathrm{\infty }]$. We have $$\int f_n\mathsf{d}\mu \le \int g\mathsf{d}\mu$$ for all $n\in \mathbb{N}$ by monotonicity. Therefore $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu \le \int \underset{r\to \mathrm{\infty }}{lim}{f}_r\mathsf{d}\mu$$ holds. It remains to prove the reverse inequality.

For the reverse inequality fix $0<\alpha <1$ and fix $0\le \varphi \le g$ a simple function. Put $$E(n)=\{x\in X:f_n(x)\ge \alpha \varphi (x)\}$$ and note that $\varphi \le f$ together with $0<\alpha <1$ implies the sequence $n\mapsto E_n$ is increasing and has union $X$. Writing $\{a_1,\dots ,a_r\}$ for the range if $\varphi$ we can then calculate that $$\begin{array}{rl}\int 1_{E(n)}\varphi \mathsf{d}\mu & =\sum _{i=1}^r{a}_r\mu ({\varphi }^{-1}(a_r)\cap E(n))\\ & \to \sum _{i=1}^r{a}_r\mu ({\varphi }^{-1}(a_r))=\int \varphi \mathsf{d}\mu \end{array}$$ where the convergence follows from continuity of measures. But then $$\underset{n\to \mathrm{\infty }}{lim}\int f_n\mathsf{d}\mu \ge \underset{n\to \mathrm{\infty }}{lim}\int 1_{E(n)}{f}_n\mathsf{d}\mu \ge \underset{n\to \mathrm{\infty }}{lim}\int 1_{E(n)}\alpha \varphi \mathsf{d}\mu =\alpha \int \varphi \mathsf{d}\mu$$ which finishes the proof because $0<\alpha <1$ was arbitrary and $0\le \varphi \le f$ was arbitrary. ▮

As a consequence of the monotone convergence theorem we get linearity and a useful convergence result.

Theorem

Whenever $f,g:X\to [0,\mathrm{\infty }]$ are measurable $$\int f+g\mathsf{d}\mu =\int f\mathsf{d}\mu +\int g\mathsf{d}\mu$$ holds.

Proof:

Fix sequences $n\mapsto {\varphi }_n$ and $n\mapsto {\psi }_n$ of simple functions that take non-negative values and satisfy ${\varphi }_n\to f$ and ${\psi }_n\to g$. Then $n\mapsto {\varphi }_n+{\psi }_n$ is a sequence of non-negative simple functions that converges to $f+g$. We have already checked that $$\int {\varphi }_n+{\psi }_n\mathsf{d}\mu =\int {\varphi }_n\mathsf{d}\mu +\int {\psi }_n\mathsf{d}\mu$$ and the monotone convergence theorem allows us to take the limits on both sides by evaluating the limits inside the integrals, giving the desired conclusion. ▮