Lecture 16

Recap

Recap from the previous.
$L_1(\mu )=\left\{\text{measure}f:\underset{\mu }{\underbrace{X}}\to \mathbb{C}|\underset{={\sup }_{0\le s\le |f|}\int sd\mu }{\underbrace{\int |f|d\mu }}<\infty \right\}$
$s={\sum }_{i=1}^n{a}_i\times X_{A_i}$
$\int sd\mu ={\sum }_{i=1}^n{a}_i\times \mu (A_i)$

$f=\mathrm{Re}f+i\mathrm{Im}f=(\mathrm{Re}f{)}^{+}-(\mathrm{Re}f{)}^{-}+i((\mathrm{Im}f{)}^{+}-(\mathrm{Im}f{)}^{-})$
$0\le \int (\mathrm{Re}f{)}^{\pm }d\mu \le \int |f|d\mu$

Proposition

$L^1(\mu )$ is a vector space under pointwise operations, and the $\int :L^1(\mu )\to \mathbb{C}d\mu$ is linear, and $|\int fd\mu |\le \int |f|d\mu$.

Proof

Let $f,g\in L^1(\mu )$, $a\in \mathbb{C}$.

Assume first that $f,g\ge 0$ simple with distinct values $a_i$ and $b_i$, let $A_i=\{x\in X|f(x)=a_i\}$ and $B_j=\{x\in X|g(x)=b_j\}$

Then ${\int }_{A_i\cap B_j}(f+g)d\mu =(a_i+b_j)\times \mu (A_i\cap B_j)$
$=a_i\times \mu (A_i\cap B_j)+b_j\times \mu (A_i+B_j)$
$={\int }_{A_i+B_j}fd\mu +{\int }_{A_i+B_j}gd\mu$

Note

$X\underset{\text{disjoint union}}{\underbrace{=}}\cup (A_i\cap B_j)$
$A\mapsto {\int }_{A}sd\mu$
Measure $L1$.

$⟹{\int }_X(f+g)d\mu =\int fd\mu +\int gd\mu$
LHS $={\sum }_{ij}\int (f+g)d\mu ={\sum }_{ij}\left({\int }_{A_i\cap B_j}fd\mu +{\int }_{A_i\cap B_j}gd\mu \right)$
$={\sum }_{ij}{\int }_{A_i\cap B_j}fd\mu +{\sum }_{ij}{\int }_{A_i\cap B_j}gd\mu$
${=}^{L_1}{\int }_{X}fd\mu +{\int }_{X}gd\mu$

$L2$:
$0\le s_1\le s_2\le \cdots \le f$
$f={\lim }_{n\to \infty }{s}_n$

By $L_2$ and Lebesgue’s Monotone Convergence Theorem (LMCT) $⟹\int$ is additive for all non-negative measurable functions $f$ and $g$.
$\int (f+g)d\mu {=}^{L_2}\int {\lim }_{n\to \infty }(s_n+s_n^{\prime })d\mu {=}^{\text{LMCT}}{\lim }_{n\to \infty }\int (s_n+s_n^{\prime })d\mu$
(from above) $={\lim }_{n\to \infty }(\int s_{n}d\mu +\int s_n^{\prime }d\mu$
($\lim$ splits) ${\lim }_{n\to \infty }\int s_{n}d\mu +{\lim }_{n\to \infty }\int s_n^{\prime }d\mu$
${=}^{\text{LMCT}}\int fd\mu +\int gd\mu$

For the complex case we have $\int \underset{\le |a|\times |f|+|g|}{\underbrace{|af+g|}}d\mu \le \int (|a|\times |f|+|g|)d\mu$
$=\int |a|\times |f|d\mu +\int |g|d\mu =|a|\times \underset{<\infty }{\underbrace{\int |f|}}d\mu +\underset{<\infty }{\underbrace{\int |g|d\mu }}<\infty$, so $af+g\in L^1(\mu )$ when $f,g\in L^1(\mu )$.
So $L^1(\mu )$ is a vector space.

Why it's $af+g$ and not $af+bg$

Usually you would use $af+bg$ and have to prove for the $bg$ part as well, but $af+g$ is enough for the proof as you can do:
$af+bg=af+(bg+0)$
as the parts in the bracket would belong in $L1$ as well.

Then use the definition of the integral for complex functions to see that it is additive. Clearly $\int afd\mu =a\int fd\mu$ when $a\ge 0$. For $a<0$ use $(-g{)}^{\pm }=g^{\mp }$ etc. Check for $a=i$, but this is okay as $\int ifd\mu =\int i(\mathrm{Re}f+i\mathrm{Im}f)d\mu =\int (-\mathrm{Im}f+i\mathrm{Re}f)d\mu =-\int \mathrm{Im}fd\mu +i\int \mathrm{Re}fd\mu =i\times \int fd\mu$.
So $\int afd\mu =a\int fd\mu ,\forall a\in \mathbb{C}$, and the integral is a linear function.

Finally, pick $b\in \mathbb{C}$ such that $|\int fd\mu |=b\times \underset{z}{\underbrace{\int fd\mu }}$

What is happening at $z$

Remember the rules of complex numbers:
$|z|=b\times z$
$b=\frac{|z|}{z}$

(linear) $=\int bfd\mu =\int \underset{\le |bf|}{\underbrace{\mathrm{Re}(bf)}}d\mu \le \int |bf|d\mu =\int \underset{=1}{\underbrace{|b|}}\times |f|d\mu =\int |f|d\mu$
QED.

Lebesgue’s Dominated Convergence Theorem

It is a theorem for complex valued functions.
Also see the definition.

Let $\{f_n\}$ be measurable functions $f_n:X\to \mathbb{C}$, and $\mu$ measure on $X$. Assume $\exists g\in L^1$ such that all $f_n\le g$.
Then ${\lim }_{n\to \infty }\int f_{n}d\mu =\int \underset{f}{\underbrace{\underset{n\to \infty }{\lim }{f}_n}}d\mu$.
If you have a measure so that the whole space is finite, $\mu (x)<\infty$:
$⟹\int 1d\mu =\mu (x)<\infty$ (can use $g$ instead of the $1$)

Proof

By Fatou’s Lemma
$\int 2gd\mu \underset{f=\lim f_n}{\underbrace{=}}\int {\lim }_{n\to \infty }\inf (\underset{\ge 0}{\underbrace{2g-|f_n-f|}})d\mu$
(by Fatou’s Lemma) $\le {\lim }_{n\to \infty }\inf \int (2g-|f_n-f|)d\mu$
(by previous proposition) $\int 2gd\mu +{\lim }_{n\to \infty }\inf \left(-\int |f_n-f|d\mu \right)$
$=\int 2gd\mu -{\lim }_{n\to \infty }\sup \int |f_n-f|d\mu$ so ${\lim }_{n\to \infty }\sup \int |f_n-fd\mu \le 0$, and ${\lim }_{n\to \infty }\underset{\ge 0}{\underbrace{\int |f_n-f|d\mu }}\le 0$, so ${\lim }_{n\to \infty }\int |f_n-f|d\mu =0$.

Then ${\lim }_{n\to \infty }|\int f_{n}d\mu -\int fd\mu |$
(by previous proposition) $={\lim }_{n\to \infty }|\int (f_n-f)d\mu |$
(again, by previous proposition) $\le {\lim }_{n\to \infty }\int |f_n-fd\mu =0$.

QED.

Another example

${\lim }_{n\to \infty }{\lim }_{m\to \infty }\frac{n}{m}={\lim }_{n\to \infty }0=0$
Then swap the limits
${\lim }_{m\to \infty }{\lim }_{n\to \infty }\frac{n}{m}={\lim }_{m\to \infty }\infty =\infty$
Which are two different numbers as $0\ne \infty$.