Lecture 15

Lebesgue’s Monotone Convergence Theorem

Say $X$ has a measure $\mu$, and let $f_n:X\to [0,\infty ]$ be measurable and $f_1\le f_2\le f_3\le \dots$.

Then $\int f_{m}d\mu \to \int {\lim }_{n\to \infty }{f}_{n}d\mu$ as $m\to \infty$.

Note

Left Integral:
$\int fd\mu ={\sup }_{0\le s\le f}\int sd\mu$

Right Integral:
${\lim }_{m\to \infty }\int f_{m}d\mu =\int {\lim }_{m\to \infty }{f}_{m}d\mu$

Proof

Note that $f\equiv {\lim }_{n\to \infty }{f}_n:X\to [0,\infty ]$ is a measurable function as

Note

• $\equiv$ is pointwise
• To make $[0,\infty ]$, you can do "$[0,a⟩,⟨a,b⟩,⟨b,\infty ]$"

$f^{-1}([0,a⟩)={\cap }_{n=1}^{\infty }\underset{\text{Measurable}}{\underbrace{f_n^{-1}([0,a⟩)}}$

More on the right, measurable, function

$x\in X$ such that
$f(x)<a$
$⟹f_n(x)\le f(x)<a\forall n$
$f_n(x)<a\forall n$
$⟹f(x)<a$

Let $b\equiv {\lim }_{n\to \infty }\int f_{n}d\mu \le \int fd\mu$ as $f_n\le f$.
Let $0\le s\le f$, $s$ measurable simple function, and $c\in ⟨0,1⟩$.
Let $A_n=\{x\in X|c\times s(x)\le f_n(x)\}=(\underset{measurablefunction}{\underbrace{f_n-cs}}{)}^{-1}(\underset{open}{\underbrace{[0,\infty ]}})$.

Then $A_1\subset A_2\subset A_3\subset \dots$ measurable, and ${\cup }_n{A}_n\stackrel{\text{(*)}}{\overbrace{=}}X$

Continuing (*)

Say $x\in X$.
If $f(x)=0$, then $x\in A_1$.
If $f(x)>0$, then $c\times s(x)<f(x)$, so $c\times s(x)<f_n(x)$ for some $n$, and $x\in A_n$.

By the previous two lemmas, we have
$b\ge {\lim }_{n\to \infty }{\int }_{A_n}{f}_{n}d\mu \ge {\lim }_{n\to \infty }c\times {\int }_{A_n}sd\mu$

Note on the $A_n$

${\int }_A\subset {\int }_X$
$A\subset X$

$=c{\lim }_{n\to \infty }{\int }_{A_n}sd\mu \stackrel{\text{2 lemmas}}{\overbrace{=}}c\times {\int }_{\cup A_n}sd\mu =c\times \int sd\mu$, so $b\ge c\times \int fd\mu$ and $b\ge \int fd\mu$

Reminder of the two lemmas

1. $A\mapsto {\int }_{A}sd\mu$ measure ($s=1⟹{\int }_{A}sd\mu =\mu (A)$)
2. For any measure $\nu$ and $A_1\subset A_2\subset \dots$ measurable $⟹\nu (\cup A_n)={\lim }_{n\to \infty }\nu (A_n)$

QED.

Corollary - Fatou’s Lemma

Defined in the lecture here: Fatou’s Lemma

Definition

Have measure $\mu$ on $X$, and $f_n:X\to [0,\infty ]$ measurable. Then $\int {\lim }_{n\to \infty }\inf f_{n}d\mu \le {\lim }_{n\to \infty }\inf \int f_{n}d\mu$

Proof

Use Lebesgue’s Monotone Convergence Theorem on $g_m={\inf }_{n\ge m}{f}_n$.
$g_1\le g_2\le \dots$ are measurable functions.
QED

Lebesgue’s Dominated Convergence Theorem

(Also defined here, it’s the same thing)

Let $g$ be a real function on $X$.

Define $g^{+}=\max \{g,0\}$, $g^{-}=-\min \{g,0\}$.

Then $g=g^{+}-g^{-}$ and $g^{\pm }\ge 0$.

Example

Definition

Given measure $\mu$ on $X$.
Define $L^{\prime }(\mu )=\left\{f:X\to \mathbb{C}\text{measurable and}\int |f|d\mu <\infty \right\}$.

Define integral for $f\in L^{\prime }(\mu )$ by $\int fd\mu \equiv \int (\mathrm{Re}f{)}^{+}d\mu -\int (\mathrm{Re}f{)}^{-}d\mu +i\int (\mathrm{Im}f{)}^{+}d\mu -i\int (\mathrm{Im}f{)}^{-}d\mu$.

Use $f=\mathrm{Re}f+i\mathrm{Im}f=(\mathrm{Re}f{)}^{+}-(\mathrm{Re}f{)}^{-}+i((\mathrm{Im}f{)}^{+}-(\mathrm{Im}f{)}^{-})$.

The integral definition makes sense as each integral on the RHS is finite.
($(\mathrm{Re}f{)}^{+}\le |f|$)

Lemma

Given measure $f:X\to [0,\infty ]$.

Then $\exists$ measurable simple functions $s_n$ such that

1. $0\le s_1\le s_2\le \cdots \le f$
2. ${\lim }_{n\to \infty }{s}_n=f$ pointwise

Proof

Define $h_n:[0,\infty ]\to [0,\infty ⟩$ by

Continue like this.

Have $0\le h_1\le h_2\le \cdots \le h_n\to l\text{as}n\to \infty$.

Set $s_n=h_n\circ f$.

---

Exercise part of the session.

These exercises are from Exercise 8.

Question 3

Prove $\mu (A)\le \mu (B)$ when $A\subset B$

Proof

Have $B=A\cup (\underset{B\setminus A}{\underbrace{A^{\complement }\cap B}})$, so

What this set looks like

$\mu (B)=\mu (A)+\underset{\ge 0}{\underbrace{\mu (B\setminus A)}}$
$⟹\mu (B)>+\mu (A)$.

Question 4

Show that if $X$ has a $\sigma$-algebra and $f:X\to Y$ set. Then the collection $N$ of subsets $A\subset Y$ such that $f^{-1}(A)$ measurable, is a $\sigma$-algebra.

$N\equiv \{A\subset Y|f^{-1}(A)\in M\}$
prove that $N$ is a $\sigma$-algebra.

Proof

1. Have $\varnothing \in N$ since $f^{-1}(\varnothing )=\varnothing \in M$.
2. If $A\in N$, then $f^{-1}(A)\in M$, so $f^{-1}(A{)}^{\complement }\in M=f^{-1}(A^{\complement })⟹A^{\complement }\in N$.
3. If $A_n\in N$, then $f^{-1}(A_n\in M)$, so $f^{-1}(\cup A_n)=\cup f^{-1}(A_n)\in M$, so $\cup A_n\in N$.

A question I don’t know the number of

Say $f:X\to Y$ and they are topological spaces.
Show that if $f$ is measurable, then $f^{-1}(A)$ is Borel measurable for any Borel set $A\subset Y$.

Proof

Consider $N=\{A\subset Y|f^{-1}(A)\text{Borel}\}$. This is a $\sigma$-algebra.

Example

It contains all the open sets in $Y$ since $f$ is measurable and then $f^{-1}(V)$ is Borel measurable for $V$ open.

Hence $N$ contains all the Borel sets in $Y$. If $A$ is Borel, then $A\in N$, so $f^{-1}(A)$ is Borel. (Note: not sure if on this if the “Borel”s are about them being Borel sets or Borel measurable)

Question 5

$X$ with $\sigma$-algebra $M$.
$f:X\to [0,\infty ]$ is measurable $⟺f^{-1}(⟨a,\infty ])\in M,\forall a>0$.

Proof

$\Rightarrow$

“Obvious”.

$\Leftarrow$

In $[0,\infty ]$ balls are $[0,a⟩$, $⟨a,\infty ]$, $[0,\infty ]$, or $⟨a,b⟩$ for $a,b\in \mathbb{R}$.

Any open set in $[0,\infty ]$ is a countable union of these “building blocks”.

Checking they belong to $M$:

• $[0,a]$
• $[0,b⟩\cap ⟨a,\infty =⟨a,b⟩$
• $[0,a⟩={\cup }_{n=1}^{\infty }\left[0,a-\frac{1}{n}\right]$
  QED.

$f_n:X\to [0,\infty ]$ is measurable
$f\equiv \sup f_n$ is measurable

Can also use infimum instead

$\inf f_n=-\sup (-f_n)$

Now need to check if $f^{-1}(⟨a,\infty ])$ is measurable and is in $M$

$f^{-1}(⟨a,\infty ])=\{x\in X|f(x)>a\}$
$={\cup }_n{f}_n^{-1}(⟨a,\infty ])\in M$
Why is the set and the union both the same?
${\cup }_n\{x\in X|f_n(x)>a\}=\{x\in X|f_n(x)>a\forall n\}$