# Thursday October 17

## Review of Tonelli

**Theorem (Tonelli)**:
Suppose $f(x, y)$ is non-negative on $\RR^{n_1} \cross \RR^{n_2}$ and measurable on the product space.
Then for a.e. $x$, we have

1. $f_x(y)$ is a measurable function of $y$.

2. Since it's non-negative and measurable, the integral makes sense, and $$
\int_{\RR^{n_2}} f(x, y) ~dy
$$ 
is a measurable function of $x$.

3. 
$$
\int\int f(x,y) ~dx ~dy = \int\int f(x,y) ~dy~dx = \int f
.$$

*Proof:*

> Qual problem alert: useful technique!

Essentially use Fubini, and truncate domain/range with a $k\dash$ball.

See proof of the *case-jumping lemma* and notes on webpage for details.

> We'll never need to dig into the proof of this, but there will *always* be a question related to *applying* it.

## Measurability of Linear Transformations

**Theorem:**
Let $T \in \GL(n, \RR)$. Then

- $f$ measurable $\implies$ $f\circ T$ is measurable. 
  - Contrast to what happens for $g$ a continuous function instead of $T$.

- 
$$
f \leq 0 \text{ or } f\in L^1 \implies \int f = \abs{\det(T)} \int (f\circ T)(x)
.$$

- 
$$
E \in \mathcal{M}(\RR^n) \implies T(E) \in \mathcal{M}(\RR^n)
.$$

It suffices to prove this for Borel sets and Borel measurable functions.
This can be proved using Fubini.

Note that if we choose $f$ to be Borel measurable, then $f\circ T$ will be measurable because $T$ is continuous.

> This follows because $\theset{E \suchthat T\inv(E) \in\mathcal{B}}$ is in fact a $\sigma\dash$algebra that contains all open sets.
> *Exercise*: Prove this.

We can also reduce this to proving the result for $T_i$ an elementary matrix, since if it holds for $T,S$ then it holds for $TS$ because
$$
\int f = \det T \int f\circ T = \det T \det S \int f\circ T \circ S = \det(TS) \int f\circ(TS)
$$
But this follows from Fubini-Tonelli.

> Note that if (3) holds for Borel sets, then (3) holds for Lebesgue null sets.

Suppose now that $f$ is just Lebesgue measurable, and let $G$ be open in $\RR$.

Then $f\inv(G) = H \union N$ where $H\in G_\delta$ and $m(N) = 0$ and 
$$
T\inv(f\inv(G)) = T\inv(H) \union T\inv(N)
.$$ 

But by the first part, $T\inv(H)$ is still Borel, and $T\inv(N)$ is still null.
So $f\circ T$ is measurable.

> Note that this kind of thing usually works -- just establish something Borel sets, then use this characterization to extend it to Lebesgue.