# Thursday September 12th

## Measurable Functions

Let $E\subseteq \RR^n$ be measurable. 
Then $f: E \to \RR \union \theset{\pm \infty}$ is *Lebesgue measurable* iff 
$$
\theset{x \in E \suchthat f(x) > a} = f\inv((a, \infty])) \in \mathcal M
,$$ 
the collection of Lebesgue measurable sets, for every $a\in\RR$.

Similarly, $E$ is Borel measurable if we replace $\mathcal M$ by $\mathcal B$, the collection of all Borel measurable sets.

![Image](figures/2019-09-10-11:11.png)\

As usual, there are many different characterizations:

- $f\inv((a, \infty])) \in \mathcal M \quad\forall a\in \RR$, since we can write this as $\intersect f\inv((a - \frac 1 k, \infty])$, which are all measurable.
- $f\inv([a, \infty]) \in \mathcal M \quad\forall a\in \RR$, by taking complements
- $f\inv([-\infty, a]) \in \mathcal M \quad\forall a\in \RR$.

**Theorem:**
If $f: E\to \RR$ is finite-valued, then 
$$
f \text{ is measurable } 
\iff 
f\inv(G) \in \mathcal M \text{ for all open } G\subseteq \RR
.$$

*Proof:*

$\impliedby$: 
Easy, since $(a, \infty)$ is always an open set.

$\implies$: 
Suppose $f$ is measurable and let $G\subseteq \RR$ be open, then $G = \disjoint I_i$ where each $I_i$ is an interval. Then $f\inv(G) = \union f\inv(I_i)$.

But if $I_i = (a, b)$, then $f\inv(I_i)= f\inv((a, \infty)) \intersect f\inv((-\infty, b))$, both of which are measurable by definition.

## Extending the Class of Measurable Functions

**Corollary:**
Continuous functions are in fact Borel measurable, and in particular Lebesgue measurable.

**Proposition:**
If $f: E\to \RR$ is measurable and $\varphi: \RR\to \RR$ is continuous, then $\varphi \circ f$ is measurable.

> Note: This condition can not be relaxed to just $\varphi$ being measurable. However, this does work if you require that $\varphi$ is Borel measurable.

*Proof:*
Let $G \subseteq \RR$ be open. 
Then 
$$
(\varphi \circ f)\inv (G) = f\inv(\varphi\inv(G)) \in \mathcal M
.$$ 

But $\varphi\inv(G)$ is open since $\varphi$ is continuous, and thus measurable, and since $f$ is a measurable function, this pulls back to a measurable set, so the composition is measurable.

Consequences of this proposition:
If $f$ is measurable, then so is $\abs f, \abs{f}^p, f^2, e^{cf}$, for any constant, etc.

We will show that $\mathcal M$ is closed under most algebraic and limiting operations.

**Theorem 2:**
If $f, g$ are $\RR\dash$valued measurable functions, then $fg$ and $f+g$ are measurable.

> Note that we already know this for $g$ a constant, since $x \mapsto x+c$ and $x\mapsto cx$ are continuous functions.

*Proof:*
To come later.

An interesting consequence: if $f,g$ are measurable then $\max(f, g)$ is as well, since
$$
\max(f, g) = \frac{(f + g) + \abs{f-g}}{2}.
$$
**Theorem:**
If $\theset{f_n}$ is a sequence of $\overline{\RR}\dash$valued measurable functions, then

1. $\sup_n f_n$ and $\inf_n f_n$ are measurable
2. $\limsup_n f_n \definedas \inf_n \sup_{k\geq n} f_k$ is measurable, as is the $\liminf_n f_n$.

> Note: As a consequence, if $f_n \to f$ pointwise and each $f_n$ is measurable, then $\limsup f_n = \lim f_n = f$ is measurable.

*Proof of Theorem 2:*

Suppose $f,g$ are measurable and finite-valued.
Let $a\in \RR$ and consider 
$$
S = \theset{x \suchthat f(x) + g(x) > a }
.$$

Then 
$$
S = \theset{ x \suchthat f(x) > -g(x) + a }
,$$ 
where $f$ is measurable and so is $-g + a$.

With the following lemma, we'll be done:

**Lemma:**
If $f, h$ are measurable, then $S = \theset{x \suchthat f(x) > h(x) }$ is always a measurable set.

*Proof:*
Since $f(x) > h(x)$, there is a rational $q$ such that $h(x) < q < f(x)$.

But then 
$$
S = \union_{q\in\QQ} \theset{x \suchthat f(x) > q > h(x)} = \theset{f > q} \intersect \theset{h < q}
,$$ 
which is an intersection of measurable sets.
$\qed$

> For the set equality, just check that $x\in S \implies x \in \union\text{stuff}$, and $x\not\in S \implies x \not \in \union \text{stuff}$.

> Note: we can write $fg = \frac{(f+g)^2 - (f-g)^2}{4}$.

*Proof of Theorem 3:*

Since $\inf_n f_n(x) = - \sup f_n(x)$, we only need to show that $\sup f_n$ is measurable.

For any given $a$, we want to show that $S = \theset{x \suchthat \sup_n f_n(x) > a}$ is measurable.

Then there is some $n$ for which $f_n(x) >  a$, and the claim is that 
$$
S = \union_n \theset{x \suchthat f_n(x) > a}
.$$

But this follows formally by just checking set inclusions. 
So $S$ is a countable union of measurable sets and thus measurable.
$\qed$

## Almost Everywhere Equality

**Definition:**
If $f,g: E \to \overline \RR, \CC$ then $f=g$ *almost everywhere* (or $f=g$ a.e.) iff 
$$
\theset{x \suchthat f(x) \neq g(x)} \text{ is null }
.$$

**Fact:**
If $f$ is measurable and $f=g$ a.e., then $g$ is measurable.

This follows because 
$$
\theset{g > a} = \theset{f > g} \union Z,$$ 
where $Z \subseteq \theset{f\neq g}$, which is null, forcing $Z$ to be null as well.

**Fact:**
If $\theset{f_n}$ is a sequence of measurable functions and $f_n \to f$ pointwise a.e., then $f$ is measurable.

Note that we've replaced open, continuous, uniform etc with a new notion of "measurable" in all instances.

This new notion isn't so far from the original ones, but allows much more to be done.

**Littlewood's Principles:**

- Every measurable set is *nearly open* (See *Lebesgue density*)

- Every measurable function is *nearly continuous* (See *Lusin's Theorem*)

- Every convergent sequence of measurable functions is *nearly uniformly convergent* (See *Egorov's Theorem*)