# Tuesday September 17th? 

## Approximation by Simple Functions

**Definition:**
Let $E \subseteq \RR^n$ be measurable. Then the *characteristic function* of $E$ is defined as

\begin{align*}
\chi_E(x) \definedas 
\begin{cases} 
1 & x \in E \\ 
0 & \text{else.} 
\end{cases}
\end{align*}

**Definition:**
A *step function* is a function of the form 
$$
S(x) = \sum_{i=1}^N a_i \chi_R(x)
$$ 
where $R$ is some rectangle.

**Definition:**
A *simple function* is a function of the form 
$$
s(x) = \sum_{i=1}^N a_j \chi_{E_j}(x)
$$ 
where each $E_j$ is measurable.

**Theorem 1:**
If $f: E \to [0, \infty]$ is a non-negative measurable function, then there exists a sequence of simple functions $\theset{s_k}$ such that 
$$
s_k(x) \leq s_{k+1}(x) ~~\forall x, k \quad \text{ and } \quad 
\lim_{k\to\infty} s_k(x) = f(x) ~\forall x
.$$

**Corollary:** 
This in fact holds for $f: E \to \overline \RR$ taking on extended real values, not just positive functions.

*Proof:* 
Write $f = f^+ - f^-$, where $f^+(x) = \max\theset{f(x), 0}$.

**Theorem 2:**
If $f: E \to \overline \RR$ is measurable, there exists a sequence $\psi_k$ of *step* functions such that $\psi_k \to f$ almost everywhere.

*Proof:*
See homework 3, problem 1c.

## Lebesgue Density Theorem

**Theorem (Lebesgue Density):**
If $E\subseteq \RR^n$ is measurable, then

\begin{align*}
\lim_{r\to 0^+} \frac{m(E \intersect  B(r, x))}{m(B(r, x))} = 1 \quad  \text{for almost every } x \in E. 
\end{align*} 

## Egorov's Theorem

**Theorem (Egorov):**
Let $E \subseteq \RR^n$ be measurable with $m(E) > 0$. 

Let $f_k: E \to \RR$ be a sequence of measurable functions such that $f(x) \definedas \lim_{k\to\infty} f_k(x)$ exists a.e. and is finite valued. 

Then the convergence is *almost uniform*, i.e.

\begin{align*}
\forall\varepsilon > 0, ~\exists F \subseteq E ~\text{closed}~ \suchthat & 
m(E\setminus F) < \varepsilon ~\text{and}~ f_k \uniformlyconverges  f ~\text{on}~ F
.\end{align*}

Are these conditions really necessary?

1. If $E = \RR$, let 
$$
f_k(x)  = \indic{\abs x > k}
.$$ 
Then $f_k \to 0$ a.e. but not "almost uniformly".

2. If $E = [0, 1]$, let 
$$
f_k(x) = k\indic{0 \leq x \leq 1 - \frac 1 k}
.$$
Then 
$$
f_k \to \infty \iff 0 \leq x < 1,~ 0 \iff x = 1,$$ 
but not "almost uniformly".

## Lusin's Theorem

**Theorem (Lusin):**

Suppose $f$ is measurable and finite-valued on a measurable set $E$ with $m(E) < \infty$. 

Then $\forall \varepsilon > 0,~\exists F \subseteq E$ closed such that 
$$
m(E\setminus F) < \varepsilon 
\text{ and } 
\restrictionof{f}{F} \text{ is continuous }
.$$

> This doesn't mean that the original $f$ is actually continuous on $F$, when thought of as a function on $E$ -- we restrict the universe to only F, so e.g. we can only take sequences that are subsets of $F$ when we go to check continuity.
> Example to note:
$$
f = \chi_\QQ \intersect [0, 1]
,$$ 
which is discontinuous at every point.

## Proof of Egorov
*Proof of Egorov:*
Assume wlog that $f_k \to f$ _everywhere_.

**Lemma:**
Let $E, \theset{f_n}$ and $f$ be as in Egorov's theorem. 

Then for all $\varepsilon > 0, \alpha > 0$ there exists a closed set $F \subseteq E$ and some $k_0 \in \NN$ such that

- $m(E\setminus F) < \varepsilon$, and

- $\abs{f_k(x) - f(x)} < \alpha\quad \forall x\in F,~ k \geq k_0$.

So let $\varepsilon > 0$, then the lemma tells us that for every $j$ we can find  a closed set $F_j \subseteq E$ with 
$$
m(E\setminus F_j) < \varepsilon / 2^j
$$ 

and $k_j$ such that 

$$
\abs{f_k(x) - f(x)} < \frac 1 j \text{ on } F_j \quad \forall k \geq k_j
.$$

So take $F \definedas \intersect F_j$, which is _closed_. 

By subadditivity, we have 
$$
m(E/F) \leq \sum m(E\setminus F_j) < \varepsilon
$$ 
by construction.

Note that the convergence is *uniform*, since $k_j$ in the lemma already provided the uniform threshold for all points in $F_j$, and $F \subseteq F_j$ for every $j$. 
$\qed$

*Proof of lemma:*
Fix $\varepsilon, \alpha$. 

Define 

\begin{align*}
E_j &\definedas \theset{x\in E \suchthat~ \abs{f_k(x) - f(x)} < \alpha~~\forall k > j} \\
&= \intersect_{k=j+1}^\infty \theset{x\in E \suchthat~ \abs{f_k(x) - f(x)} < \alpha}
.\end{align*}


Note $E_j \subseteq E_{j+1}$ and $E = \union E_j$, so we have $E_j \nearrow E$.

Using continuity from below, $\lim_j m(E_j) = m(E)$.

Since $m(E) < \infty$, there exists a $k_0$ such that 
$$
j \geq k_0 \implies m(E\setminus E_j) < \varepsilon / 2
.$$

So choose $F \subset E_{k_0}$ be closed with the property that 
$$
m(E_{k_0} \setminus F) \leq \frac \varepsilon  2
.$$

So if $x\in F$, then $x\in E_{k_0}$ and thus $x\in E_j$ for all $j\geq k_0$ since they are nested.

But then 
$$
k \leq k_0 \implies \abs{f_k(x) - f(x)} < \alpha
,$$ 
and we're done. 
$\qed$

## Convergence in Measure

**Definition:**
Let $E \subseteq \RR^n$ be measurable and $f, \theset{f_k}$ be measurable, finite-valued functions defined on $E$.

We say that $f_k \to^m f$ or $f_k \to f$ *in measure* if for every $\alpha > 0$, we have 
$$
\lim_{k\to\infty} m(\theset{x \in E \suchthat \abs{f_k(x) - f(x)} > \alpha}) = 0
.$$
$\qed$

How does this relate to pointwise convergence?

**Theorem**:
If $m(E) < \infty$, then 
$$
f_k \to f ~a.e \text{ on } E \iff f_k \to^m f \text{ on } E
.$$

*Proof:* 
Exercise using the previous lemma.

> Note that the converse is false! 
See homework exercise. 
There is a partial converse: convergence in measure will yield a *subsequence* that converges almost everywhere.