# Measure Theory 2

Time for some more measure theory. Given a measure space ${(X,\mathcal M,\mu)}$, let ${L^+}$ denote the set of measurable nonnegative real-valued functions.

Theorem 1 $\mbox{}$

1. Let ${f,g\in L^+}$. Then ${\int (f+g)\mathsf d\mu = \int f \mathsf d\mu + \int g\mathsf d\mu}$.
2. Let ${f\in L^+}$ and ${c\in[0,\infty)}$. Then ${\int cf\mathsf d\mu = c\int f\mathsf d\mu}$.
3. Let ${f,g\in L^+}$ with ${f\leqslant g}$. Then ${\int f\mathsf d\mu \leqslant \int g\mathsf d\mu}$.

Proof: $\mbox{}$

1. If ${\varphi = \sum_{i=1}^n a_i\chi_{E_i}}$ and ${\psi = \sum_{j=1}^m b_j\chi_{F_j}}$ where we may assume WLOG

$\displaystyle \bigcup_{i=1}^n E_i = \bigcup_{j=1}^m F_j = X,$

then

$\displaystyle \varphi = \sum_{i=1}^n a_i\chi_{E_i\cap\bigcup_{j=1}^m F_j}=\sum_{i=1}^n\sum_{j=1}^m a_i\chi_{E_i\cap F_j}$

and similarly

$\displaystyle \psi = \sum_{j=1}^m\sum_{i=1}^n b_j \chi_{F_j\cap E_i}.$

Hence

$\displaystyle \varphi + \psi = \sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)\chi_{E_i\cap F_j},$

so that

\displaystyle \begin{aligned} \int (\varphi+\psi)\mathsf d\mu &= \sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)\mu(E_i\cap F_j) \\ &= \sum_{i=1}^n\sum_{j=1}^m a_i\mu(E_i\cap F_j) + \sum_{i=1}^n\sum_{j=1}^m b_j\mu(E_i\cap F_j)\\ &= \int \varphi \mathsf d\mu + \int\psi \mathsf d\mu. \end{aligned}

From this it follows that if ${\varphi_n\uparrow f}$ and ${\psi_n\uparrow g}$, then ${(\varphi + \psi)_n\uparrow f+g}$ and hence by monotone convergence (to be proved later!)

\displaystyle \begin{aligned} \int (f+g)\mathsf d\mu &= \int \lim_{n\rightarrow\infty} (\varphi_n + \psi_n)\mathsf d\mu\\ &= \lim_{n\rightarrow\infty} \int (\varphi_n + \psi_n)\mathsf d\mu\\ &= \lim_{n\rightarrow\infty} \int \varphi_n \mathsf d\mu + \int \psi_n\mathsf d\mu\\ &= \int f\mathsf d\mu + \int g\mathsf d\mu. \end{aligned}

2. If ${\varphi=\sum_{i=1}^n a_i\chi_{E_i}}$ then

\displaystyle \begin{aligned}\int c\varphi\mathsf d\mu &= \sum_{i=1}^n ca_i\mu(E_i)\\ &= c\sum_{i=1}^n a_i\mu(E_i)\\ &= c\int \varphi\mathsf d\mu.\end{aligned}

Hence if ${\varphi_n\uparrow f}$, we have ${c\varphi_n\uparrow cf}$, and thus

$\displaystyle \int cf\mathsf d\mu = \lim_{n\rightarrow\infty} \int c\varphi_n\mathsf d\mu = \lim_{n\rightarrow\infty} c\int\varphi_n\mathsf d\mu = c\int f\mathsf d\mu.$

3. Since ${g-f\geqslant 0}$, this follows immediately from

$\displaystyle 0\leqslant \int (g-f)\mathsf d\mu = \int g\mathsf d\mu - \int f\mathsf d\mu.$

$\Box$

One remark: For ${f\in L^+}$ and ${E\in\mathcal M}$ we define

$\displaystyle \int_E f\mathsf d\mu = \int f\chi_E\mathsf d\mu.$

It follows from monotonicity that if ${E,F\in\mathcal M}$ with ${E\subset F}$, then for ${f\in L^+}$, ${f\chi_E\leqslant f\chi_F}$ and hence

$\displaystyle \int_E f\mathsf d\mu = \int f\chi_E\mathsf d\mu \leqslant \int f\chi_F\mathsf d\mu = \int_F f\mathsf d\mu.$

Now for the monotone convergence theorem:

Theorem 2 Let ${(X,\mathcal M,\mu)}$ be a measure space and ${\{f_n\}}$ a sequence of measurable real-valued functions satisfying ${f_n\geqslant 0}$, ${f_n\leqslant f_{n+1}}$ and ${f_n\rightarrow f}$ pointwise. Then ${f}$ is measurable and

$\displaystyle \int f\mathsf d\mu = \lim_{n\rightarrow\infty}\int f_n\mathsf d\mu.$

Proof: ${\int f_n\mathsf d\mu}$ is a sequence of nondecreasing real numbers, and thus has a limit (possibly ${+\infty}$). Since ${f_n\leqslant f}$ for each ${n}$, we have ${\int f_n\mathsf d\mu\leqslant \int f\mathsf d\mu}$ and thus

$\displaystyle \lim_{n\rightarrow\infty} \int f_n\mathsf d\mu \leqslant \int f\mathsf d\mu.$

For the reverse inequality, let ${\alpha\in(0,1)}$ and let ${\varphi}$ be a simple function with ${\varphi\leqslant f}$. Set ${E_n=\{x\in X: f_n(x)\geqslant\alpha\varphi(x) \}}$. Since ${f_n\rightarrow f}$ and ${\alpha\varphi\leqslant f}$, we have ${\bigcup_{n=1}^\infty E_n = X}$. Hence for each ${n}$,

$\displaystyle \int f\mathsf d\mu \geqslant \int_{E_n} f_n\mathsf d\mu \geqslant \alpha\int_{E_n}\varphi\mathsf d\mu.$

Letting ${n\rightarrow\infty}$ we obtain

$\displaystyle \lim_{n\rightarrow\infty}\int f_n\mathsf d\mu \geqslant \alpha\int \varphi\mathsf d\mu.$

Since this is true for all ${0<\alpha<1}$, it remains true for ${\alpha=1}$, and as ${\varphi}$ was arbitrary, taking the supremum over ${\varphi\leqslant f}$ we have

$\displaystyle \lim_{n\rightarrow\infty}\int f_n\mathsf d\mu \geqslant \int f\mathsf d\mu.$

$\Box$