# Thursday October 3

## Relating Zero Functions to Zero Integrals Over Measurable Sets
**Theorem:**
$f = 0$ a.e. iff $\int_E f = 0$ for all $E\in\mathcal{M}$.

If $f \in L^+$ we already know that $f=0$ a.e. iff $\int f = 0$.

$\implies$: 
Since $f = 0$ a.e., we have $\abs f = 0$ a.e. and since $\abs{f} \in L^+$ we have $\int \abs{f} = 0$.

Now let $E \in \mathcal{M}$; then

\begin{align*}
\abs{\int_E f} \leq \int_E \abs{f} \leq \int \abs{f} = 0
.\end{align*}

$\impliedby$:
Suppose $\int_E f = 0 ~~\forall E \in \mathcal{M}$ and $f\neq 0$ a.e., then either

1. $f+$ is positive on a set of nonzero measure, or

2. $f^-$ is positive on a set of nonzero measure.

So suppose wlog (1) holds.

Let $E = \theset{x \suchthat f^+ > 0}$, then $m(E) > 0$.

Then $\int_E f^+ > 0$, since $\chi_E f^+ \neq 0$ almost everywhere.

We also know that $f^+ \in L^+$, so 
$$
f^+ = 0 \text{ a.e. } \iff \int f = 0
.$$
But then $\int _E f > 0$, since 
$$
\mathrm{support}(f^+) \intersect \mathrm{support}(f^-) = \theset{x \suchthat f(x) = 0}
,$$ 
so $f^- = 0$ on $E$. 

$\qed$

## Approximation Theorems and Dense Subspaces of $L^1$

**Definition:**
We say that a collection $\mathcal{C}$ of functions is *dense* in $L^1$ iff
$$
\forall\varepsilon>0 ~\text{ and }~ \forall f\in L^1,
\quad \exists g\in\mathcal{C} 
\quad \text{such that} \quad \norm{f-g}_1 < \varepsilon.
$$

**Theorem(s):**

1. Simple functions are dense in $L^1$. **(DCT)**

2. Continuous functions with compact support ($C_c$ or $C_0$) are dense in $L^1$.

3. Step functions are dense in $L^1$.

*Proof of (1):*
Let $f\in L^1$ and $\varepsilon > 0$.

Since $f$ is measurable, there exists a sequence of simple functions $\theset{\phi_k} \to f$ pointwise with $\abs{\phi_k} \leq \abs{\phi_{k+1}}$.
Then $f$ dominates $\phi_k$ and the DCT yields $\int\abs{\phi_k - f} < \varepsilon$ for $k$ large enough.

$\qed$

> We'll use this as a stepping stone -- we really want to get *continuous functions*, but now we can show there are continuous functions arbitrarily close to *simple* functions, and the triangle inequality will give us the desired result.

*Proof of (2):*
We have shown that there exists a simple function $\phi = \sum_{j=1}^N a_j \chi_{E_j}$ in standard representation, where $a_j \neq 0$ and the $E_j$ are disjoint, with $\int \abs{f - \phi} < \varepsilon$.

It suffices to show that for all $j$, there exists a $g_j \in C_c$ such that $\norm{\chi_{E_j} - g_j} < \varepsilon$.

Note that if we have this, we can define $g = \sum a_j g_j \in C_c$. 

But then

\begin{align*}
\int\abs{\phi - g} = \int \abs{\sum_i^N a_i (\chi_{E_j} - g_j) } \leq \sum_i^N \abs{a_i} \abs{\chi_{E_j} - g_j} \leq C\varepsilon
.\end{align*}


Then applying the triangle inequality yields the desired result.

**Important Observation:**

Each $E_j$ has finite measure, so we have 
$$
m(E_j) = \frac{1}{\abs{a_j}} \int_{E_j} \abs{\phi} \leq \frac{1}{\abs{a_j}}\int \abs{\phi} < \infty
.$$

**Claim:**
If $m(E) < \infty$, then there exists a $g\in C_c$ such that $\norm{\chi_E - g}_1 < \varepsilon$ for any $\varepsilon > 0$.

*Proof:*
Note that we can find a $K \subseteq E \subseteq G$ such that $K$ is compact, $G$ is open, and $m(G\setminus K) < \varepsilon$.

Since $K$ is closed and $G^c$ is closed, **by Urysohn's Lemma**, there is a continuous $g$ such that $\chi_K \leq g \leq \chi_G$.
But then $g$ is zero on $G^c$ and 1 on $K$.

Then $\abs{\chi_E - g}$ is supported on $G\setminus K$, so 
$$
\int\abs{\chi_E - g} \leq m(G\setminus K) < \varepsilon
.$$ 
$\qed$

> Remark: We will eventually show that *smooth* compactly supported functions are also dense in $L^1$.

This approximation theorem yields some nice proofs:

## Small Tails and Absolute Continuity

**Proposition:**
If $f\in L^1$ and $\varepsilon > 0$, then 

1. **Small tails**: 
$$
\exists N \text{ such that } \int_{\norm{x} \geq N} \abs{f} < \varepsilon
$$ 
  *Take $f_N = f\chi_{B(N)} \nearrow f$*

2. **Absolute continuity:** 
There exists a $\delta > 0$ such that 
$$
m(E) < \delta \implies \int_E \abs{F} < \varepsilon
$$ 
  *Take $f_N = f\chi_S$ where $S = \theset{f(x) \leq N}$.*

> **Useful technique:** If you want to prove something for $L^1$ functions, try to show it's true for $C_c$ functions.

Note that we know $\exists g\in C_c$ such that $\int \abs{f-g} < \varepsilon$.

*Proof of (1):*
Let $N$ be large enough such that $g=0$ if $\abs x \geq N$.
Let $E = \theset{x \suchthat \abs x \geq N}$.

Then
$$
\int_E \abs{f} = \int_E \abs{f - g + g} \leq \int_E\abs{f-g} + \int_E\abs{g} < \varepsilon  + 0.
$$

*Proof of (2):*
There exists an $M$ such that $\abs{g} \leq M$, since $C_c$ functions are bounded almost everywhere.

Then 
$$
\int_E \abs g \leq M \cdot m(E) < \varepsilon.
$$

So set $\delta = \varepsilon/M$, then if $m(E) < \delta$ then
$$
\int_E \abs f \leq \int\abs{f-g} + \int_E\abs{g} < \varepsilon.
$$

## Continuity in $L^1$

> **Qual problem alert:** Prove the following theorem. 
Note that DCT doesn't quite work!

**Theorem (Continuity in $L^1$)**
$$
f\in L^1 \implies \lim_{h\to 0} \int \abs{f(x+h) - f(x)} = 0
.$$

*Proof:*
Let $\varepsilon > 0$. 

Then choose $g$ such that 
$$
\int\abs{f(x) - g(x)} < \varepsilon
.$$

By translation invariance, $\int\abs{f(x+h)- g(x+h)} < \varepsilon$ as well.

> Qual problem alert: remember how to prove translation invariance of the Lebesgue integral.

Now 
$$
\int\abs{f(x+h) - f(x)} \leq 2\varepsilon + \int\abs{g(x+h) - g(x)}
.$$

Since $g$ is continuous and has compact support, $g$ is uniformly continuous.

So enlarge the support of $g$ to a compact set $K$ such that $\abs{g(x+h) - g(x)} = 0$ for all $x\in K^c$ and $\abs h \leq 1$.
But then 
$$
\int_K \abs{g(x+h) - g(x)} \leq \varepsilon \int_K 1 \to 0
.$$

> Note that $\mathrm{supp}(F) = \overline{\theset{x \suchthat f(x) \neq 0}}$.