# Measure Theory 1

Let ${X}$ be a nonempty set. A collection of subsets ${\mathcal M\subset 2^X}$ is said to be a ${\sigma}$-algebra if

• ${\varnothing\in\mathcal M}$
• ${E\in\mathcal M}$ implies ${E^c\in\mathcal M}$
• If ${\{E_n\}_{n=1}^\infty}$ is a countable sequence of sets in ${\mathcal M}$, then ${\bigcup_{n=1}^\infty E_n\in\mathcal M}$.

If ${\mathcal E\subset 2^X}$, the ${\sigma}$-algebra generated by ${\mathcal E}$ is the intersection of all ${\sigma}$-algebras ${\mathcal F}$ containing ${\mathcal E}$:

$\displaystyle \sigma(\mathcal E) = \bigcap_{F\supset E}\mathcal F.$

If ${X}$ is a nonempty set and ${\mathcal M}$ is a ${\sigma}$-algebra, the pair ${(X,\mathcal M)}$ is called a measurable space. A function ${\mu:\mathcal M\rightarrow[0,\infty]}$ is said to be a measure if the following conditions hold:

• ${\mu(\varnothing)=0}$
• If ${\{E_n\}_{n=1}^\infty}$ is a countable sequence of disjoint sets in ${\mathcal M}$, then ${\mu\left(\bigcup_{n=1}^\infty E_n \right)=\sum_{n=1}^\infty \mu(E_n)}$.

If ${(X,\mathcal M)}$ is a measurable space and ${\mu}$ is a measure defined on ${\mathcal M}$, the triple ${(X,\mathcal M,\mu)}$ is called a measure space.

Theorem 1 If ${E,F\in\mathcal M}$ and ${E\subset F}$, then ${\mu(E)\leqslant \mu(F)}$.

Proof:

$\displaystyle \mu(F) = \mu((F\cap E)\cup(F\cap E^c)) = \mu(F\cap E) + \mu(F\cap E^c) = \mu(E) + \mu(F\cap E^c)\geqslant \mu(E).$

$\Box$

Theorem 2 If ${E_1, E_2,\ldots\in\mathcal M}$ then

$\displaystyle \mu\left(\bigcup_{n=1}^\infty E_n \right)\leqslant \sum_{n=1}^\infty \mu(E_n).$

Proof: Put ${F_1 = \varnothing}$ and ${F_n = E_n\setminus E_{n-1}}$ for ${n=2,3,\ldots}$. Then for any ${N}$, ${\bigcup_{n=1}^N F_n = \bigcup_{n=1}^N E_n = E_N}$, and for any ${i,j}$, ${F_i\cap F_j=\varnothing}$. Hence

$\displaystyle \mu\left(\bigcup_{n=1}^\infty E_n \right) = \mu\left(\bigcup_{n=1}^\infty F_n \right) = \sum_{n=1}^\infty \mu(F_n)\leqslant \sum_{n=1}^\infty \mu(E_n),$

as ${F_n\subset E_n}$. $\Box$

Theorem 3 If ${E_1, E_2, \ldots\in\mathcal M}$ and ${E_n\subset E_{n+1}}$ for ${n=1,2,\ldots}$ then

$\displaystyle \mu\left(\bigcup_{n=1}^\infty E_n \right)=\lim_{n\rightarrow\infty}\mu(E_n).$

Proof: Put ${F_1 = \varnothing}$ and ${F_n = E_n\setminus E_{n-1}}$ for ${n=2,3,\ldots}$. Then for any ${N}$, ${\bigcup_{n=1}^N F_n = \bigcup_{n=1}^N E_n = E_N}$, and for any ${i,j}$, ${F_i\cap F_j=\varnothing}$. Hence

\displaystyle \begin{aligned} \mu\left(\bigcup_{n=1}^\infty E_n \right) &= \mu\left(\bigcup_{n=1}^\infty F_n \right)\\ &= \sum_{n=1}^\infty \mu(F_n)\\ &= \lim_{N\rightarrow\infty}\sum_{n=1}^N \mu(F_n)\\ &=\lim_{N\rightarrow\infty}\mu\left(\bigcup_{n=1}^N F_n \right)\\ &= \lim_{N\rightarrow\infty} \mu(E_N). \end{aligned}

$\Box$

Let ${(X,\mathcal M,\mu)}$ be a measure space, and ${(Y,\mathcal N)}$ be a measurable space. A function ${f:X\rightarrow Y}$ is said to be measurable if for all ${E\in\mathcal N}$, ${f^{-1}(E)\in\mathcal M}$.

Theorem 4 If ${\mathcal N=\sigma(\mathcal E)}$, then ${f:X\rightarrow Y}$ is measurable if and only if ${f^{-1}(E)\in\mathcal M}$ for all ${E\in\mathcal E}$.

Proof: The set ${\{E\subset \mathcal N: f^{-1}(E)\in\mathcal M \}}$ is a ${\sigma}$-algebra that contains ${\mathcal M}$. $\Box$

The Borel ${\sigma}$-algebra on ${\mathbb R}$ is the ${\sigma}$-algebra generated by the open sets. If ${E\in\mathcal M}$, the characteristic function of ${E}$ is the function ${\chi_E: X\rightarrow\{0,1\}}$ defined by

$\displaystyle \chi_E(x) = \begin{cases}1,& x\in E\\ 0,& x\notin E.\end{cases}$

A simple function is a function of the form

$\displaystyle \varphi = \sum_{i=1}^n a_i\chi_{E_i},$

where ${a_i\geqslant 0}$ and ${E_i\in\mathcal M}$. For a simple function ${\varphi}$, define the Lebesgue integral by

$\displaystyle \int_{\Omega}\varphi\,\mathsf d\mu = \sum_{i=1}^n a_i\mu(E_i).$

Theorem 5 If ${f:X\rightarrow[0,\infty)}$ is Borel-measurable, there exists a sequence of simple functions ${\{\varphi_n\}}$ such that ${\varphi_n\leqslant\varphi_{n+1}}$ and ${\varphi_n}$ converges pointwise to ${f}$.

Proof: For each nonnegative integer ${n}$ and each nonnegative integer ${k<2^{2n}}$, define

\displaystyle \begin{aligned} E_{n,k} &= \left\{x\in X: \frac{k}{2^n}< f(x)\leqslant \frac{k+1}{2^n} \right\}\\ F_n &= \{x\in X: 2^n < f(x)\leqslant 2^{n+1} \}. \end{aligned}

Define

$\displaystyle \varphi_n = \sum_{k=0}^{2^{2n}-1}\frac k{2^n}\chi_{E_{n,k}} + 2^n\chi_{F_n}.$

Observe that for each ${n}$,

$\displaystyle \{x\in X: f(x)>0 \} = \left(\bigcup_{k=0}^{2^{2n}-1}E_{n,k}\right)\cup\left(\bigcup_{k=n}^\infty F_n \right),$

and this union is disjoint. If ${x\in E_{n,k}}$ then

$\displaystyle \frac{k2^n}{2^{n+1}} < f(x)\leqslant \frac{(k+1)2^n}{2^{n+1}},$

so that

$\displaystyle \varphi_n(x) = \frac k{2^n} = \frac{2k}{2^{n+1}}\leqslant \varphi_{n+1}(x).$

If ${x\in F_n\cap F_{n+1}^c}$ then

$\displaystyle \frac k{2^{n+1}}

for some ${2^{2n+1}\leqslant k <2^{2n+2}}$, so that

$\displaystyle \varphi_n(x) = 2^n = \frac{2^n2^{n+1}}{2^{n+1}}=\frac{2^{2n+1}}{2^{n+1}}\leqslant \frac k{2^{n+1}}=\varphi_{n+1}(x).$

If ${x\in F_n\cap F_{n+1}}$, then

$\displaystyle \varphi_n(x) = 2^n < 2^{n+1} = \varphi_{n+1}(x).$

If ${f(x)=\infty}$, then ${\lim_{n\rightarrow\infty}\phi_n(x)=\lim_{n\rightarrow\infty}2^n=\infty}$. Otherwise, for sufficiently large ${n}$, ${0\leqslant f(x)-\varphi_n(x)\leqslant\frac1{2^n}}$, from which it follows that ${\varphi_n\rightarrow f}$. $\Box$

For ${f:X\rightarrow[0,\infty]}$, define the integral of ${f}$ by the supremum of simple functions less than ${f}$:

$\displaystyle \int_X f\,\mathsf d\mu = \sup_{\varphi\leqslant f}\int_X\varphi\,\mathsf d\mu.$

For ${f:X\rightarrow[-\infty,\infty]}$ define the positive and negative parts of ${f}$ by

\displaystyle \begin{aligned} f^+ &= \max\{f,0\}\\ f^- &= \max\{-f, 0\} \end{aligned}

Then ${f=f^+-f^-}$ where ${f^+\geqslant 0}$ and ${f^-\geqslant0}$, so we can define the integral of ${f}$ by

$\displaystyle \int_X f\,\mathsf d\mu = \int_X f^+\,\mathsf d\mu - \int_X f^-\,\mathsf d\mu.$

For ${f:X\rightarrow\mathbb C}$, the integral is defined as

$\displaystyle \int_X f\,\mathsf d\mu = \int_X (\mathrm{Re} f)^+\mathsf d\mu - \int_X(\mathrm{Re}f)^-\mathsf d\mu + \left(\int_X(\mathrm{Im} f)^+\mathsf d\mu - \int_X(\mathrm{Im} f)^-\mathsf d\mu \right)i.$