# Tuesday September 24th 

## Convergence Theorems

Two main convergence theorems: define $L^+ = \theset{f: \RR^n \to [0, \infty] \suchthat f\text{ is measurable.}}$.
Then

**Theorem 1 (MCT):**
If $\theset{f_n} \subseteq L^+$ with $f_n \nearrow f$, then 
$$
\int f = \lim \int f_n.$$

**Theorem 2 (Fatou's lemma):**
If $\theset{f_n} \subseteq L^+$, then 
$$
\int \liminf f_n \leq \liminf \int f_n.
$$

**Corollary 1:**
If $\theset{f_n} \subseteq L^+$ and $f_n \to f$ a.e. and $\int f_n \leq M$ uniformly, then 
$$
\int f \leq M
$$ 

> This uses Fatou's lemma.


**Corollary 2:**
If $\theset{f_n} \subseteq L^+$ and $f_n \to f$ a.e. with instead $f_n \leq f$ a.e. for all $n$, then 
$$
\int f = \lim \int f_n
.$$

*Proof:*
By Fatou, 
$$
\int f \leq \liminf \int f_n
.$$

If we can show $\limsup \int f_n \leq \int f$ as well, we're done.

Since integrals satisfy monotonicity, 
$$
f_n \leq f \implies \int f_n \leq \int f ~\text{a.e.}
$$ 
But by order-limit laws, we then have $\limsup \int f_n \leq \int f$ as well. 
$\qed$

## Extending the Integral to $\overline \RR\dash$valued functions (and $\CC$)

**Definition:**
A function $f: \RR^n \to \overline \RR$ is *integrable* iff $\int \abs f < \infty$.
Note that $\abs f = f_+ + f_-$, so if $f:\overline \RR \to \RR$ is integrable, then 
$$
\int \abs f = \int (f_+ + f_-) = \int f_+ + \int f_-,
$$ 
which means both must be finite.
We now have two finite *numbers*, so we can subtract.

**Definition:**
For $f:\RR^n \to \overline \RR$, we define 
$$
\int f = \int f_+ - \int f_-.
$$
Similarly, for $f: \CC \to \overline \RR$, let $f = \mathrm{Im} f + i\mathrm{Re} f$, and define
$$
\int f = \int \Re f + i\int \Im f.
$$

> Note: the space of all $\overline \RR$ or $\CC$ valued functions forms a real (resp. complex) vector space, and the integral is a real (complex) linear functional. 
This is not immediate -- multiplying by scalars is clear, but distributing integrals over sums is not. 
If $h = f + g$, then it is **not** the case that $h_+ = f_+ + g_+$. But we can write out 
$$
h = f + g \implies h_+ + h_- = f_+ + f_- + g_+ + g_-
,$$ 
maneuver things so that everything is positive, and *then* take integrals.

## $L^1$ and its Norm
**Definition**:
We can provisionally define 
$$
L^1 = \theset{f: \RR^n \to \CC: f\text{ is measurable }},
$$

where we'd like to define 
$$
\norm{f}_{L^1} \definedas \int \abs f
.$$ 

However, this only yields a *seminorm*, since nonzero functions still end up with zero norm.
This can be remedied by identifying functions which agree on a set of measure zero.

**Proposition (Triangle Inequality for $L^1$ Seminorm):**
If $f\in L^1$, then $\abs{\int f} \leq \int \abs f$.

*Proof:*
This is trivial if $\int f = 0$.
If $f$ is $\RR\dash$valued, then

\begin{align*}
\abs{\int f} 
&= \abs{\int f_+ - \int f_-} \\
&\leq \abs{\int f_+} + \abs{\int f_-} \\
&= \int f_+ + \int f_- \\
&= \int f_+ + f_- \\
&= \int \abs{f}
.\end{align*}

If $f$ is $\CC\dash$valued, then $\abs z = \frac{z^* z}{\abs z}$, so

\begin{align*}
\frac{(\int f)^*}{\abs{\int f}} \int f 
&\definedas \alpha \int f \\
&= \int \alpha f \\
&= \int \mathrm{Re}(\alpha f) + i\int\mathrm{Im}(\alpha f) 
.\end{align*}

but since what we started with was *real*, the imaginary component vanishes, so this equals
$$
\int \mathrm{Re}(\alpha f) = \abs{\int \mathrm{Re}(\alpha f) } \leq \int \abs{\mathrm{Re}(\alpha f) } \leq \int \abs{\alpha f} = \int \abs{f}.
$$

> Note: this is referred to as a **rotation/triangle trick**.


**Actual definition of $L^1$:**
Let $X \subseteq \RR^n$ be measurable. 
Then
$$
L^1(X) = \theset{\text{equivalence classes of a.e. defined integrable functions on $X$}}
$$

This is an equivalence relation, and we write $f\sim g \iff f = g$ a.e.

We then define $\norm{f}_1 = \int_X \abs f$.

We have 
$$
\int \abs f = 0 ~\text{a.e.} \iff f = 0 ~\text{a.e.}
$$

*Exercise*: Prove this for $L^+$ functions.


We'd like this to also be true iff $\int_E f = 0$ for all measurable $E \subseteq X$.

## Dominated Convergence Theorem

The following theorem will apply whenever we want to switch integrals and limits, but $f$ is *not necessarily* in $L^+$.

> This is the ONLY theorem that doesn't require non-negativity!!!

**Theorem (DCT):**
Suppose $\theset{f_n} \in L^1$, $f_n \to f$ a.e., and $\abs{f_n} \leq g$ a.e. for all $n$ with $g\in L^1$.

Then $f\in L^1$ and 
$$
\int f = \lim \int f_n \quad\text{and}\quad \lim \int\abs{f_n - f} = 0
$$

> Note that the second statement is stronger, and in fact implies the first. This statement is what we'll prove.

*Proof:*
Since $f_n \to f$ and each $f_n$ is measurable, then $f$ is measurable.
Since $f_n \leq g$ for all $n$, then $f \leq g$.
So 
$$\int \abs{f} \leq \int \abs{g} < \infty
,$$ 
and thus $f\in L^1$.

It suffices to show that $\limsup \int \abs{f_n - f} \leq 0$.

We have to get something non-negative to apply anything we know so far, so

\begin{align*}
0 \leq \abs{f_n - f} \leq \abs {f_n} + \abs f \leq g + \abs f \\
\implies g + \abs f - \abs{f_n - f} \geq 0 \\
(\text{This is where we'll apply Fatou.}) \\
\implies \int \liminf (g + \abs f - \abs{f_n -f}) &\leq \liminf \int (g + \abs f - \abs{f_n - f})\\
\implies \int (g + \abs f) &\leq \liminf \int(g + \abs f) - \liminf \int\abs{f_n - f} \\
\implies \int (g + \abs f) &\leq \int(g + \abs f) + \limsup \int\abs{f_n - f} \\
\implies 0 &\leq \limsup \int\abs{f_n - f}. \qed
\end{align*}

$\qed$

## Differentiating Under the Integral

Let
$$
F(t) = \int f(x, t)~dx
$$

- Is $F$ continuous at a point $t_0$?
- Is $F$ differentiable at $t_0$?

We could show continuity by looking at 

\begin{align*}
\lim_{t \to t_0} \abs{F(t) - F(t_0)} 
\leq \lim_{t\to t_0} \int \abs{f(x, t) - f(x, t_0)} 
\leq_{DCT} \int \lim \abs{f(x, t) - f(x, t_0)}
.\end{align*}

which will go to zero exactly when $f$ is continuous in $t$.


Differentiability can be shown by considering

\begin{align*}
\lim \frac{\abs{F(t_0) - F(t)}}{t - t_0}
&\leq \lim \int \frac{f(x, t) - f(x, t_0)}{t - t_0} ~dx \\
&\leq \int \lim \frac{f(x, t) - f(x, t_0)}{t - t_0}~dx \\
&= \int f'(x, t_0)~dx
.\end{align*}