# Tuesday, October 8

Notation: think of $\RR^n = \RR^{n_1} \cross \RR^{n_2}$ where $n_1 + n_2 = n$.

**Motivation**: 
If $f(x,y)$ is measurable, is it true that $f(y) \definedas f(x_0, y)$ also measurable for a fixed $x_0$?

## Fubini and Tonelli

**Theorem (Tonelli):**
Let $f(x, y)$ be non-negative and measurable on $\RR^n$.
Then for almost every $x\in \RR^{n_1}$, we have

1. $$f_x(y) \definedas f(x, y)$$ is measurable as a function of $y$ in $\RR^{n_2}$

2. $$F(x) \definedas \int f(x, y) ~dy$$ is measurable as a function of $x$,

3. $$G(y) = \int F(x) ~dx = \int \left( \int f(x, y) ~dy\right) ~dx$$ is measurable and equal to $\int_{\RR^n} f$.

**Corollary:**
If $E \subset \mathcal{M}~(\RR^{n_1} \cross \RR^{n_2})$, then for a.e. $x\in \RR^{n_1}$, the slice 
$$
E_x \definedas \theset{y\in\RR^{n_2} \suchthat (x,y) \in E}
$$
is measurable.

Moreover, $x \mapsto m(E_x)$ is a measurable function of $x$, and 
$$
m(E) = \int_{\RR^n} m(E_x) ~dx
.$$

**Warning:** 
We assumed $E$ was measurable here, but it is possible for every slice to be measurable while $E$ itself is not!

Take $E = \mathcal{N} \cross I$ for $\mathcal{N}$ the unmeasurable set. 
Then $E_x = \chi_{[0, 1]}$ and so the image is always measurable.

But taking $y$ slices yields $E_y \chi_{\mathcal{N}}$, which (by the above corollary) would have to be measurable if $E$ were measurable.

![Image](figures/2019-10-08-11:35.png)\

> Note: We need to show that taking a **cylinder** on a function (i.e. given $f(x)$ and defining $F(x, y) = f(x)$) does not destroy measurability. 
This is necessary in the context of convolution, since $f(x - y)$ will need to be measurable in both variables in order to apply Tonelli.

## Application: Area Under the Graph

Suppose $f \geq 0$ on $\RR^n$, with no assumption of measurability.

Consider defining the "area under the graph" as 
$$
\mathcal{A} \definedas \theset{(x, y) \in\RR^n \cross \RR \suchthat 0 \leq y \leq f(x)}
.$$

Then

1. $f$ is measurable on $\RR^n$ iff $\mathcal{A}$ is a measurable subset of $\RR^{n+1}$.

2. If $f$ is measurable on $\RR^n$, then 
$$
m(\mathcal{A}) = \int_{\RR^n} f = \int_0^\infty m(\theset{x\in\RR^n \suchthat f(x) \geq y}) ~dy
$$

![Image](figures/2019-10-08-11:33.png)\

*Proof of (1):*

$\implies:$ Suppose $f$ is measurable on $\RR^n$.

By the lemma, $F(x, y) = f(x)$ is measurable on $\RR^n \cross \RR$ and $G(x, y) = y$ is as well.
But then $\mathcal{A} = \theset{G \leq F} \intersect \theset{G \geq 0}$, which is an intersection of measurable sets and thus measurable.


$\impliedby:$ 
Suppose $\mathcal{A}$ is measurable.

By Tonelli, for almost every $x\in \RR^n$, the slice 
$$
\mathcal{A}_x = \theset{y \in \RR \suchthat (x, y) \in \mathcal{A}} = [0, f(x)]
$$
is measurable.

Then $m(\mathcal{A}_x) = f(x)$, so $x\mapsto \mathcal{A}_x$ is a measurable function of $x$ and $m(\mathcal{A})  = \int f(x) ~dx$.

Repeating this argument with $y$ slices instead, for almost every $y\in \RR$ we have 
$$
\mathcal{A}_y = \theset{x \in \RR^n \suchthat (x, y) \in \mathcal{A}} = \theset{x\in\RR^n \suchthat f(x) \geq y \geq 0}
,$$ 
which is a measurable subset of $\RR^n$.

So it makes sense to integrate it, and 
$$
m(\mathcal{A}) 
= \int m(\mathcal{A}_y)~dy 
= \int_0^y m(\theset{x \in \RR^n \suchthat f(x) \geq y}) ~dy
.$$

$\qed$

*Alternative proof*:

\begin{align*}
\int_0^\infty m ( \theset{x\in \RR^n \suchthat f(x) \geq y}) 
&= \int_{\RR} \int_{\RR^n} \chi_{S \definedas \theset{x\in \RR^n \suchthat f(x) \geq y \geq 0}} \\
&= \int_{\RR^n} \int_\RR \chi_S \\
&= \int_{\RR^n} \int_0^{f(x)} ~dy~dx \\
&= \int_{\RR^n} f(x)~dx
.\end{align*}

$\qed$

## Appendix on Measurability in $\RR^{n_1} \cross \RR^{n_2}$.

**Lemma:**
If $f$ is measurable on $\RR^{n_1}$, then $F(x, y) \definedas f(x)$ is measurable on the product space $\RR^{n_1} \cross \RR^{n_2}$

> Qual problem alert.

*Proof of Lemma:*
Suppose $f$ is measurable on $\RR^n$; we want to show that $F(x, y) = f(x)$ is measurable on $\RR^n \cross \RR$.

This amounts to showing that for any $a$, 
$$
S_a \definedas \theset{(x,y) \suchthat F(x, y) \geq a} \in \mathcal{M}(\RR^{n+1}).$$

But we can rewrite 
$$
S_a = \theset{x\in\RR^n \suchthat f(x) > a} \cross \RR
,$$ 

which is the cylinder on a measurable set. 
As we will show, this is always measurable.

> Best way to show measurability: use Borel characterization, or show that it's an $H \disjoint N$ where $H \in F_\sigma$ and $N$ is null.

So write $E = H \disjoint N$ where $H\in F_\sigma$ and $N$ is null.

Then $E \cross \RR = (H \cross \RR) \union (N \cross \RR)$.

But $H\cross \RR$ is still an $F_\sigma$ set, so we just need to show $N\cross \RR$ is still null.

We have $N \cross [-k, k] \nearrow N \cross \RR$, so we can use continuity from below.

To see that $m(N \cross [-k, k]) = 0$, first cover $N$ by such that $\sum \abs{Q_i} < \varepsilon / 2k$.

But the measure of any rectangle over such a cube will be $M(\overline{Q}_i) = 2 k \cdot m(Q_i)$, which we can pull out of $\sum \abs{\overline{Q}_i}$.
$\qed$

## Fubini and Fubini-Tonelli

**Summary**":

- **Tonelli**: *Non-negative and measurable* allows switching integrals, 

- **Fubini**: *Just measurable* allows switching the integrals, the integrals are finite, and all iterated variants are equal. 

- **Fubini/Tonelli**: Extends switching beyond just non-negative integrands.
  

**Theorem (Fubini):**
Let $f(x, y)$ be an integrable function on $\RR^n \cross \RR^{n_1} \cross \RR^{n_2}$.

Then for almost every $x\in \RR^{n_1}$,

1. $f_x(y) = f(x, y)$ is an *integrable* function of $y$ in $\RR^{n_2}$.

2. $\int_{\RR^{n_2}} f(x, y) ~dy$ is an integrable function of $x$ in $\RR^{n_1}$.

Moreover, 
$$
\int_{\RR^n} f = \int \int f(x, y) ~dx~dy
$$ 
in either order.

**Theorem (Fubini-Tonelli):**
Let $f(x, y)$ be measurable in the product space.

If either 
$$
\int \left( \int \abs{f(x, y) ~ dy}\right) ~dx < \infty
$$
or
$$
\int \left( \int \abs{f(x, y) ~ dx}\right) ~dy < \infty
$$
then by Tonelli on $\abs{f(x, y)}$, we can conclude $f\in L^1(\RR^{n_1} \cross \RR^{n_2})$.

Moreover, by Fubini, $\int f$ is equal to either iterated integral.

> Moral: If *any* iterated integral is finite, then they *all* are.


Comparing this to sums: recall that $\sum \int f_n = \int \sum f_n$ is true exactly when

1. $f_n \geq 0$, and
2. $\sum \int \abs{f_n} < \infty$.