Measure Theory/Integration

Let $(X,\sigma ,\mu )$ be a $\sigma$-finite measure space. Suppose $s$ is a positive simple measurable function, with $s=\sum _{i=1}^3{y}_i{\chi }_{A_i}$; $A_i\in \sigma$ are disjoint.

Define $\underset{X}{\int }sd\mu =\sum y_i\mu (A_i)$

Let $f:X\to \overline{ℝ}$ be measurable, and let $f\ge 0$.

Define $\underset{X}{\int }fd\mu =\sup \{\underset{X}{\int }sd\mu =\sum y_i\mu (A_i)|s\text{ simple },s\ge 0,s\le f\}$

Now let $f$ be any measurable function. We say that $f$ is integrable if $f^{+}$ and $f^{-}$ are integrable and if $\underset{X}{\int }{f}^{+}d\mu ,\underset{X}{\int }{f}^{-}d\mu <\mathrm{\infty }$. Then, we write

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

The class of measurable functions on $X$ is denoted by ${ℒ}^1(X)$

For $0<p<\mathrm{\infty }$, we define ${ℒ}^p$ to be the collection of all measurable functions $f$ such that $|f{|}^p\in {ℒ}^1$

A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.

Properties

Let $(X,\sigma ,\mu )$ be a measure space and let $f,g$ be measurable on $X$. Then

1. If $f\le g$, then $\underset{X}{\int }fd\mu \le \underset{X}{\int }gd\mu$
2. If $A,B\in \sigma$, $A\subset B$, then $\underset{A}{\int }fd\mu \le \underset{B}{\int }fd\mu$
3. If $f\ge 0$ and $c\ge 0$ then $\underset{X}{\int }cfd\mu =c\underset{X}{\int }fd\mu$
4. If $E\in \sigma$, $\mu (E)=0$, then $\underset{E}{\int }fd\mu =0$, even if $f(E)=\{\mathrm{\infty }\}$
5. If $E\in \sigma$, $f(E)=\{0\}$, then $\underset{E}{\int }fd\mu =0$, even if $\mu (E)=\mathrm{\infty }$

Proof

Monotone Convergence Theorem

Suppose $f_n\ge 0$ and $f_n$ are measurable for all $n$ such that

1. $f_1(x)\le f_2(x)\le \dots$ for every $x\in X$
2. $f_n(x)\to f(x)$ almost everywhere on $X$

Then, $\underset{X}{\int }{f}_{n}d\mu \to \underset{X}{\int }fd\mu$

Proof

$\underset{X}{\int }{f}_{n}d\mu$ is an increasing sequence in $ℝ$, and hence, $\underset{X}{\int }{f}_{n}d\mu \to \alpha \in \overline{ℝ}$ (say). We know that $f$ is measurable and that $f\ge f_n\mathrm{\forall }n$. That is,

$\underset{X}{\int }{f}_{1}d\mu \le \underset{X}{\int }{f}_{2}d\mu \le \dots \underset{X}{\int }{f}_{n}d\mu \le \dots \underset{X}{\int }fd\mu$

Hence, $\alpha \le \underset{X}{\int }fd\mu =\sup \{\underset{X}{\int }sd\mu :s\text{ is simple },0\le s\le 1\}$

Let $c\in [0,1]$

Define $E_n=\{x|f_n(x)\ge cs(x)\}$; $n=1,2\dots$. Observe that $E_1\subset E_2\subset \dots$ and $\bigcup E_n=X$

Suppose $x\in X$. If $f(x)=0$ then $s(x)=0$ implying that $x\in E_1$. If $f(x)>0$, then there exists $n$ such that $f_n(x)>cs(x)$ and hence, $x\in E_n$.

Thus, $\bigcup E_n=X$, therefore $\underset{X}{\int }{f}_n(x)d\mu \ge \underset{E_n}{\int }{f}_{n}d\mu \ge c\underset{E_n}{\int }sd\mu$. As this is true if $c\in [0,1]$, we have that $\alpha \ge \underset{X}{\int }fd\mu$. Thus, $\underset{X}{\int }{f}_{n}d\mu \to \underset{X}{\int }fd\mu$.

Fatou's Lemma

Let $f_n\ge 0$ be measurable functions. Then,

$\underset{X}{\int }\mathrm{lim\; inf}{f}_{n}d\mu \le \mathrm{lim\; inf}\underset{X}{\int }{f}_{n}d\mu$

Proof

For $k=1,2,\dots$ define $g_k(x)=\underset{i\ge k}{\inf }{f}_i(x)$. Observe that $g_k$ are measurable and increasing for all $x$.

As $k\to \mathrm{\infty }$, $g_k(x)\to \underset{n\ge k}{\mathrm{lim\; inf}}{f}_n(x)$. By monotone convergence theorem,

$\underset{X}{\int }{g}_{k}d\mu \to \underset{X}{\int }\mathrm{lim\; inf}{f}_n(x)d\mu$ and as $\underset{X}{\int }{g}_k(x)d\mu \le \mathrm{lim\; inf}\int g_k(x)d\mu$, we have the result.

Dominated convergence theorem

Let $(X,\sigma ,\mu )$ be a complex measure space. Let $\{f_n\}$ be a sequence of complex measurable functions that converge pointwise to $f$; $f(x)=\underset{n\to \mathrm{\infty }}{\lim }{f}_n(x)$, with $x\in X$

Suppose $|f_n(x)|\le g(x)$ for some $g\in {ℒ}^1(X)$ then

$f\in {ℒ}^1$ and $\underset{X}{\int }|f_n-f|d\mu \to 0$ as $n\to \mathrm{\infty }$

Proof

We know that $|f|\le g$ and hence $|f_n-f|\le 2g$, that is, $0\le 2g-|f_n-f|$

Therefore, by Fatou's lemma, $\underset{X}{\int }2gd\mu \le \mathrm{lim\; inf}\underset{X}{\int }(2g-|f_n-f|)d\mu \le \underset{X}{\int }2gd\mu +\mathrm{lim\; inf}\underset{X}{\int }(-|f_n-f|)d\mu$

$=\underset{X}{\int }2gd\mu -\mathrm{lim\; sup}\underset{X}{\int }|f_n-f|d\mu$

As $g\in {ℒ}^1$, $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}\underset{X}{\int }|f_n-f|d\mu \le$ implying that $\mathrm{lim\; sup}\underset{X}{\int }|f_n-f|d\mu$

Theorem

1. Suppose $f:X\to [0,\mathrm{\infty }]$ is measurable, $E\in \sigma$ with $\mu (E)>0$ such that $\underset{E}{\int }fd\mu =0$. Then $f(x)=0$ almost everywhere $E$
2. Let $f\in {ℒ}^1(X)$ and let $\underset{E}{\int }fd\mu =0$ for every $E\in \sigma$. Then, $f=0$ almost everywhere on $X$
3. Let $f\in {ℒ}^1(X)$ and $\left|\underset{X}{\int }fd\mu \right|=\underset{X}{\int }|f|d\mu$ then there exists constant $\alpha$ such that $|f|=\alpha f$ almost everywhere on $E$

Proof

1. For each $n\in \mathbb{N}$ define $A_n=\{x\in E|f(x)>\frac{1}{n}\}$. Observe that $A_n\uparrow E$
   but $\frac{1}{n}\mu (A_n)\le \underset{A_n}{\int }fd\mu \le \underset{E}{\int }fd\mu =0$ Thus $\mu (A_n)=0$ for all $n$, by continuity, $f=0$ almost everywhere on $E$
2. Write $\underset{E}{\int }fd\mu =\underset{E}{\int }{u}^{+}d\mu -\underset{E}{\int }{u}^{-}d\mu +i(\underset{E}{\int }{v}^{+}d\mu -\underset{E}{\int }{v}^{-}d\mu )$, where $u^{+},u^{-},v^{+},v^{-}$ are non-negative real measurable.
   Further as $\underset{E}{\int }{u}^{+}d\mu ,\underset{E}{\int }(-u^{-})d\mu$ are both non-negative, each of them is zero. Thus, by applying part I, we have that $u^{+},u^{-}$ vanish almost everywhere on $E$. We can similarly show that $v^{+},v^{-}$ vanish almost everywhere on $E$.

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