Measure Theory/Integration

< Measure Theory

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

Define $\displaystyle\int_Xs~d\mu=\sum y_i\mu(A_i)$

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

Define $\displaystyle\int_Xf~d\mu=\sup\{\int_Xs~d\mu=\sum y_i\mu(A_i)|s\text{ simple },s\geq0,s\leq f\}$

Now let $f$ be any measurable function. We say that $f$ is integrable if $f^+$ and $f^-$ are integrable and if $\displaystyle\int_Xf^+~d\mu,\int_Xf^-~d\mu<\infty$ . Then, we write

$\displaystyle\int_Xf~d\mu=\int_Xf^+~d\mu-\int_Xf^-~d\mu$

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

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

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

If $f\leq g$ , then $\displaystyle\int_Xfd\mu\leq\int_Xgd\mu$
If $A,B\in\sigma$ , $A\subset B$ , then $\displaystyle\int_Afd\mu\leq\int_Bfd\mu$
If $f\geq 0$ and $c\geq 0$ then $\displaystyle\int_Xcfd\mu=c\int_Xfd\mu$
If $E\in\sigma$ , $\mu(E)=0$ , then $\displaystyle\int_Efd\mu=0$ , even if $f(E)=\{\infty\}$
If $E\in\sigma$ , $f(E)=\{0\}$ , then $\displaystyle\int_Efd\mu=0$ , even if $\mu(E)=\infty$

Proof

Monotone Convergence Theorem

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

$f_1(x)\leq f_2(x)\leq\ldots$ for every $x\in X$
$f_n(x)\to f(x)$ almost everywhere on $X$

Then, $\displaystyle\int_Xf_nd\mu\to\int_Xfd\mu$

Proof

$\displaystyle\int_Xf_nd\mu$ is an increasing sequence in $\mathbb{R}$ , and hence, $\displaystyle\int_Xf_nd\mu\to\alpha\in\overline{\mathbb{R}}$ (say). We know that $f$ is measurable and that $f\geq f_n\forall n$ . That is,

$\displaystyle\int_Xf_1d\mu\leq\int_Xf_2d\mu\leq\ldots\int_Xf_nd\mu\leq\ldots\int_Xfd\mu$

Hence, $\displaystyle\alpha\leq\int_Xfd\mu=\sup\{\int_Xsd\mu:s\text{ is simple },0\leq s\leq1\}$

Let $c\in [0,1]$

Define $E_n=\{x|f_n(x)\geq cs(x)\}$ ; $n=1,2\ldots$ . Observe that $E_1\subset E_2\subset\ldots$ 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 $\displaystyle\int_Xf_n(x)d\mu\geq\int_{E_n}f_nd\mu\geq c\int_{E_n}sd\mu$ . As this is true if $c\in [0,1]$ , we have that $\alpha\geq\displaystyle\int_Xfd\mu$ . Thus, $\displaystyle\int_Xf_nd\mu\to\int_Xfd\mu$ .

Fatou's Lemma

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

$\displaystyle\int_X\liminf f_n d\mu\leq\liminf\int_Xf_nd\mu$

Proof

For $k=1,2,\ldots$ define $g_k(x)=\displaystyle\inf_{i\geq k}f_i(x)$ . Observe that $g_k$ are measurable and increasing for all $x$ .

As $k\to\infty$ , $g_k(x)\to\displaystyle\liminf_{n\geq k}f_n(x)$ . By monotone convergence theorem,

$\displaystyle\int_Xg_kd\mu\to\int_X\liminf f_n(x)d\mu$ and as $\displaystyle\int_Xg_k(x)d\mu\leq\liminf\int g_k(x)d\mu$ , 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 $f$ ; $f(x)=\displaystyle\lim_{n\to\infty}f_n(x)$ , with $x\in X$

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

$f\in\mathcal{L}^1$ and $\displaystyle\int_X|f_n-f|d\mu\to 0$ as $n\to\infty$

Proof

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

Therefore, by Fatou's lemma, $\displaystyle\int_X2gd\mu\leq\liminf\int_X(2g-|f_n-f|)d\mu\leq\displaystyle\int_X2gd\mu+\liminf\int_X(-|f_n-f|)d\mu$

$=\displaystyle\int_X2gd\mu-\limsup\int_X|f_n-f|d\mu$

As $g\in\mathcal{L}^1$ , $\displaystyle\limsup_{n\to\infty}\int_X|f_n-f|d\mu\leq$ implying that $\displaystyle\limsup\int_X|f_n-f|d\mu$

Theorem

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

Proof

For each $n\in\mathbb{N}$ define $A_n=\{x\in E|f(x)>\frac{1}{n}\}$ . Observe that $A_n\uparrow E$
but $\frac{1}{n}\mu(A_n)\leq\displaystyle\int_{A_n}fd\mu\leq\int_Efd\mu=0$ Thus $\mu(A_n)=0$ for all $n$ , by continuity, $f=0$ almost everywhere on $E$
Write $\displaystyle\int_Efd\mu=\int_Eu^+d\mu-\int_Eu^-d\mu+i\left(\int_Ev^+d\mu-\int_Ev^-d\mu\right)$ , where $u^+,u^-,v^+,v^-$ are non-negative real measurable.
Further as $\displaystyle\int_Eu^+d\mu,\int_E(-u^-)d\mu$ 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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