# Thursday January 16th

Recall some definitions:

- $\pi\dash$class: $A, B \in \mathcal A \implies AB \in \mathcal A$

- $\lambda\dash$class:
  1. $\Omega \in \mathcal A$
  2. $A \supset B \implies A\setminus B \in \mathcal A$
  3. $A_n \nearrow A \implies A \in\mathcal A$.

We can define minimal such classes containing any nonempty class $\mathcal E$.

**Claim:**
A $\lambda\pi\dash$class is a $\sigma\dash$algebra.

> Note: this just means that it is both a $\sigma$ and a $\pi$ class. 

*Proof:*

- If $A\in \mathcal A$, then $A^c = \Omega - A \in \mathcal AA$ since $\Omega, A \in \mathcal A$ and $A \subset \Omega$ (uses $\lambda\dash$class properties).

- Then $(A \union B)^c = A^c B^c \in \mathcal A$ (using $\pi\dash$class properties).

- If $A_n \in \mathcal A$ and $A_n$ are increasing, then $A_n \nearrow A \definedas \union A_n$, which is in $\mathcal A$.

Remark:
Any $\sigma\dash$algebra is a $\lambda\pi\dash$class.

Theorem:
If $\mathcal D$ is a $\pi\dash$class, then $\sigma(\mathcal C) = \lambda(\mathcal D)$.

> Here we are taking the algebras generated by each, and $\sigma(\mathcal D)$ is the Borel $\sigma\dash$algebra.

Since $\lambda(\mathcal D)$ is minimal, $\lambda(\mathcal D) \subset \sigma(\mathcal D)$.
Thus it suffices to show $\lambda(\mathcal D)$ is a $\pi\dash$class.

> Similar to proof that $\sigma(A) = m(A)$ the monotone algebra.

Let $\mathcal G_1 = \theset{ A\suchthat AB \in \lambda(\mathcal D) ~\forall B \in \mathcal{D} }$.
Want to show that $AB \in \lambda(\mathcal{D}) \implies AB \in \lambda{\mathcal{D}}$.
We can observe that $\mcg_1 \supset \mcd$ since $\mcd$ is a $\pi\dash$class and $\lambda(\mcd) \supset \mcd$, so $\Omega \in \mcg_1$.
So choose $A,D \in \mcg_1$ with $A\supset D$.
Then $AB, AD \in \lambda(\mcd)$, so $(AB - AD) = A(B-D)$ is in $\lambda(\mcd)$.

If $A_n \in \mcd$ with $A_n \nearrow \lim A_n$, then $A_n B \in \lambda(\mcd)$ for all $B \in \mcd$.
Then $(\lim A_n)B = \lim (A_n B)$, since both are in $\lambda$.
Thus $\mcg_1 \supset \lambda(\mcd)$, and for any $A \in \lambda(\mcd)$, $B\in \mcd$, we have $AB\in \lambda(\mcd)$.

Now let $\mcg_2 = \theset{B \suchthat AB \in \lambda(\mcd) ~\forall A \in \lambda(\mcd)}$.
By the above result, $\mcg_2 \supset \mcd$.
Moreover, $\mcg_2$ is a $\lambda\dash$class, which we need to verify.
But then $\mcg_2 \supset \lambda(\mcd)$, then we're done since $B\in\lambda(\mcd) \intersect \mcg_2$ implies that for all $A \in \lambda(\mcd)$, $AB \in \lambda(\mcd)$, which is what we wanted to show.

$\qed$

Notation:
Let $(\Omega, \mca)$ be a measurable space, then any $A\in \mca$ is a *measurable* set.

Letting $(\Omega_1, \mca_1), (\Omega_2, \mca_2)$ be two measurable spaces, then define $\Omega_1 \cross \Omega_2 \definedas \theset{(\omega_1, \omega_2) \suchthat \omega_i \in \Omega_i }$.
Then define $\mca_1 \cross \mca_2 = \sigma(A_1\cross A_2 \suchthat A_i\in\mca_i)$.
Note that this is a $\pi\dash$class in the case of $\RR^n$: the intersections of rectangles are still rectangles.

For $\RR^n$, we define a function $f$ to be measurable iff $f\inv(-\infty, a) = \theset{x\suchthat f(x) < a}$ is Borel-measurable for all $a\in \RR$.
Similarly, we say $f$ is measurable here iff $f\inv(A) = \theset{f \suchthat f(x) \in A}$.
We thus say $X:\Omega_1 \to \Omega_2$ is measurable iff $X\inv(\mca_2) \subset\mca_1$ where $X\inv(\mca_i) = \theset{X\inv(A) \suchthat A\in\mca_i}$ and $X\inv(A) = \theset{\omega \suchthat X(\omega) \in A}$.

Lemma:
Given a transformation $X:\Omega_1 \to \Omega_2$ and a class $\mca$ of $\Omega_2$, we have

\begin{align*}
X\inv( \sigma_{\Omega_1}(\mca) ) = \sigma_{\Omega_1}(X\inv(\mca))
.\end{align*}

Proof:
We have

1. $X\inv( \sigma_{\Omega_2}(\mca) ) \supset X\inv(\mca)$.

2. $X\inv( \sigma_{\Omega_2}(\mca) )$ is a $\sigma\dash$algebra.

  > This follows from the fact that preimages commute with union, intersection, and set difference.

So it suffices to show that $X\inv ( \sigma_{\Omega_2}(\mca))$ is the minimal $\sigma\dash$algebra containing $X\inv(\mca)$.
I.e., we want to show that for all $\sigma\dash$algebras $\mcd \supset X\inv(\mca)$, we have $\mcd \supset X\inv( \sigma_{\Omega_2}(\mca_1)  )$.

Define $\mcg \definedas \theset{ A \suchthat A \subset \Omega_2, X\inv(A) \in \mcd }$, then $\mcg \supset \mca$ and $\mcg$ is also a $\sigma\dash$ algebra.
Thus $\mcg \supset \sigma_{\Omega_2}(\mca)$, $X\inv(\mcg) \supset X\inv(\sigma_{\Omega_2}(\mca))$, thus $\mcd \supset X\inv(\mcg) \supset X\inv(\sigma_{\Omega_2}(\mca))$.

$\qed$

> Note: we did not use measurability here.

In the case of $\RR$, we had that if $f_1, f_2$ be measurable, then $f_1 \circ f_2$ is measurable.
Similarly, if we have

\begin{align*}
X: \Omega_1 &\to \Omega_2 \\
Y: \Omega_2 &\to \Omega_3
.\end{align*}

If we define $Z = Y(X): \Omega_1 \to \Omega_3$, then if $X,Y$ are measurable then $Z$ is measurable.

Proof:

\begin{align*}
Z\inv(\mca_3) &= (Y(X))\inv (\mca_3) \\
&= X\inv (Y\inv (\mca_3)) \\
&\subset X\inv(\mca_2) \\
&\subset \mca_1
.\end{align*}

Similarly, if $\theset{f_i}$ are measurable, then the function $F(x) = [f_1(x), f_2(x), \cdotss, f_n(x)]$ is measurable.

We can also let $X_i: (\Omega, \mca) \to (\Omega_i, \mca_i)$, then the function $X = (X_1, \cdots, X_n): \Omega \to \prod X_i$ is measurable with respect to the product measure (i.e. $\sigma(\prod A_i)$).

Next time: looking at sequences of functions, showing limits are measurable, expected values.