# Thursday September 19th 

## Review of the Lebesgue Integral for $L^+$

Recall the definition of $L^+$, and the fact that for any $f\in L^+$, there is a sequence $\theset{f_n}$ of simple functions in $L^+$ such that $f_n \nearrow f$.

Given any simple function $\phi$, we defined 
$$
\int \phi = \sum_j a_j m(E_j)
$$ 
**iff** this is a standard representation for $\phi$.

We then extend to all functions in $L^+$ by defining 
$$
\int f \definedas \sup \theset{\int \phi \suchthat 0\leq \phi \leq f,~\phi\text{ simple }}
.$$

**Some properties:**

- $f\leq g \implies \int f \leq \int g$ 

  (monotonicity, easy to show)

- $\int cf = c\int f$ 

  (also easy to show)

- $\int (f + g) = \int f + \int g$ 

  (less obvious, follows from MCT)

**Theorem:**
If $\theset{f_n} \subset L^+$, then $\sum \int f_k = \int \sum f_k$. 

*Proof:*
Exercise, not too tricky.

## Proof of Monotone Convergence Theorem

**Theorem (MCT):**
If $\theset{f_n} \subset L^+$ with $f_k \leq f_{k+1}$ and $f_k \to f$, then $\lim \int f_n = \int \lim f_n = \int f$.

*Proof of MCT:*
Given any simple $\phi \in L^+$, define the set function

\begin{align*}
\mu_\phi: \mathcal M \to [0, \infty] \\
A \mapsto \int_A \phi
.\end{align*}


So if $\theset{E_k} \subset \mathcal M$ and $E_1 \subseteq E_2 \subseteq \cdots$, then $\mu_\phi(\union E_k) = \lim \mu_\phi(E_k)$.
 
> Note that 
$$
f_k \in L^+ \implies \lim f_k \in L^+ \quad \text{ and }\quad  f_k \leq f_{k+1} \implies \int f_k \leq \int f_{k+1}
,$$ 
so the limit on the LHS makes sense.

Let $f = \lim f_k$, then 
$$
\int f_k \leq \int f \text{ for all } k
,$$ 
so 
$$
\lim \int f_k \leq \int f.$$

So we need to show that $\lim \int f_k \geq \int f$.

Fix $\alpha \in (0, 1)$ and let $\phi$ be simple with $0 \leq \phi \leq f$. 
(We'll later show that the result is independent of this choice.)

Let 
$$
E_k = \theset{x \suchthat f_k(x) \geq \alpha \phi(x)}
,$$ 
which we can now say is *clearly* measurable.

Then $E_1 \subseteq E_2 \subseteq \cdots$ since the $f_k$ are increasing.
Moreover, $\union E_k = \RR^n$ (check!), since $f_k \to f, \alpha \phi < \phi \leq f$.

Then

\begin{align*}
\int f_g \geq \int_{E_k} 
\geq \alpha \int_{E_k}\phi 
\definedas \alpha \mu_\phi(E_k)
.\end{align*}

But then

\begin{align*}
\lim \int f_k \geq \lim \alpha \mu_\phi(E_k) = \alpha \mu_\phi(\RR^n) = \alpha \int \phi
\quad \forall \alpha, \forall \phi
.\end{align*}


So $\lim \int f_k \geq \int \phi$ for all $\phi$ simple with $0\leq \phi \leq f$.

Thus 
$$
\int f = \sup_\phi \int \phi \leq \lim \int f_k
,$$ 
which is what we wanted to show. 
$\qed$



### Chebyshev's Inequality

**Theorem (Chebyshev's Inequality):**
If $f\in L^+$, then 
$$m(\theset{x \suchthat f(x) \geq \alpha}) \geq \frac 1 \alpha \int f \forall \alpha
.$$

*Proof:*

![Image](figures/2019-09-19-11:46.png)\

Just note that 
$$
\alpha m(\theset{x \suchthat f(x) > \alpha}) = \int \alpha \chi_{f \geq \alpha} \leq \int f
.$$ 
$\qed$

### The Integral Detects Almost-Everywhere Equality

**Proposition:**
Suppose $f\in L^+$.
Then 
$$
\int f = 0 \implies f = 0 \text{ a.e. }
$$

> Nice trick: showing $a_n \to L$ can be done by showing $a_n - L \to 0$. 
Similarly, 
$$
\int f = \int g \iff \int (f-g) = 0 
$$

*Proof of Proposition:*

This is obvious for simple functions:

If $f = \sum a_i \chi_{E_i}$ in standard representation, then 
$$
\int f = \sum a_j m(E_j) = 0 \iff \text{ either } a_j = 0 ~\forall j, \text{ or } a_j \neq 0 \text{ and } m(E_j) = 0
.$$ 

So it only disagrees with zero on a measure zero set.

In general, suppose $f = 0$ a.e., note that 
$$
\phi \leq f = 0 \implies \phi = 0 \text{ a.e. }
$$ 
and thus $\int \phi = 0$ by the previous case.

But then 
$$
\int f = \sup_\phi \int \phi = 0 \text{ a.e. }
,$$ and we're done.

Now suppose $\int f = 0$ a.e.; we can apply Chebyshev,which says that 
$$
m(\theset{x \suchthat f(x) \geq \alpha}) = 0 
\text{ for any } 
\alpha \geq 0
.$$

But then 
\begin{align*}
m(\theset{x \suchthat f(x) > 0}) 
&= m(\union_n \theset{x\suchthat f(x) > \frac 1 n}) \\
&= \lim m(\text{stuff}) \\
&= \lim 0 \\ 
&= 0
.\end{align*}

$\qed$

> *Remark:*
In the MCT, the monotonicity is necessary, i.e. we really need $f_k \nearrow f$.

*Examples:*

- $f_k = k \chi_{[0, \frac 1 k]}$. Then $f_k \to 0$ a.e. but $\int f_k = 1$ for all $k$ while $\int f = 0$.

- $f_k = \chi_{[k, k+1]} \to 0$ a.e. (the skateboard to infinity!)

## Fatou's Theorem
Another convergence theorem, this time with virtually no hypothesis:

**Theorem (Fatou):**
If $\theset{f_k} \subset L^+$, then $\int \liminf f_k \leq \liminf \int f_k$

> How to remember: in the above examples, we had $\int \lim f_k = 0$, so we can saturate the LHS to zero to obtain an inequality of the form $a\leq b$.

*Proof of Fatou:*
We can write 
$$
\liminf_k f_k = \lim_k g_k \quad \text{ where }\quad  g_k = \inf_{n \geq k} f_n
.$$

Note that $g_k \nearrow \liminf_k f_k$, we can apply MCT.
So

\begin{align*}
\int \liminf_k f_k 
&= \int \lim_k g_k \\
&=_\text{MCT} \lim_k \int g_k \\
&= \liminf \int g_k \\
&\leq \liminf \int f_k
,\end{align*}


where we can note that 
$$
g_k \leq f_k \implies \int g_k \leq \int f_k
.$$ 
$\qed$

## Dominated Convergence Theorem

**Theorem (DCT):**
If $\theset{f_k} \subset L^+$, $f_k \to f$ a.e., **and** $f_k \leq g \in L^+$ where $\int g < \infty$, then 
$$
\int f = \int \lim f_k = \lim \int f_k
.$$