# Tuesday October 1

## Completeness of $L^1$ (Revisited)

**Last time:**
$L^1$ is complete, where we used the fact that $\RR$ is complete in the following way

**Theorem:**
\begin{align*}
\RR \text{ is complete } \iff \left( \sum \abs x_n \infty \implies \sum x_n < \infty \right)
.\end{align*}

*Proof:*

$\implies$:  
Suppose $\RR$ is complete and $\sum \abs x_n < \infty$.

Let $S_N = \sum_{i=1}^N x_n$.
Then if $N > M$, 
$$
\abs{S_N - S_M} \leq \sum_{i=M+1}^N \abs x_n \to 0
.$$

$\impliedby$: 
Suppose every absolutely convergent series is convergent.

Let $\theset{x_n}$ be Cauchy; we want to show that it is convergent as well.

> Note: we'll use the same trick as last time. 
The goal is to cook up an absolutely convergent sequence, the convergence of which will imply convergence of our original series.

Choose a subsequence 
$$
n_1 \leq n_2 \leq \cdots \quad \text{ such that }\quad  \abs{x_n - x_m} < 2^{-j} \quad \text{ if }\quad  n, m \geq n_j
.$$

Let $y_1 = x_{n_1}$ and $y_j = x_{n_j} - x_{n_{j-1}}$ for $j > 1$.

Then 
$$
x_{n_k} = \sum_{i=1}^k y_j \quad \text{and} \quad \sum_{j=1}^\infty \abs{y_j} \leq \abs y_1 + \sum_{j=2}^\infty 2^{-k} < \infty.
$$

So $\lim x_{n_k}$ exists and equals $\sum y_j$.

It follows that for $n > n_k$ and $k$ is sufficiently large, 
\begin{align*}
\abs{x_n - x} \leq \abs{x_n - x_{n_k}} + \abs{x_{n_k} - x} < \varepsilon
.\end{align*}

$\qed$

**Theorem (Modified):**
Let $X$ be a normed vector space.

$$
X \text{ is complete } \iff \left( \sum_n \norm {x_n} < \infty \implies \sum_n x_n < \infty \right)
.$$

*Proof:*
Completely the same, just replace absolute values with norms everywhere!

## Translation and Dilation Invariance of the Lebesgue Integral

> Qual Problem Alert!
 
**Definition**:
Define a *translation* $\tau_h(x) \definedas x+h$ and $\tau f(x) \definedas f(x-h)$ for all $h\in\RR\units$.

**Definition:**
Define a *dilation* $f_\delta(x) \definedas \delta^{-n} f(\delta\inv x)$ for all $\delta > 0$.

**Theorem:**

1.
\begin{align*}
f\in L^1 \implies \tau_h f\in L^1 &\text{ and } \int \tau_h f = \int f \\ 
&\left( \text{i.e. } \int_E f(x-h) = \int_{E + h} f \right)
.\end{align*}

2.
\begin{align*}
f\in L^1 \implies f_\delta \in L^1 &\text{ and } \int f_\delta = \int f \\ 
&\left( \text{i.e. } \delta^{-n} \int f(\delta\inv x) = \int f(y) \right)
.\end{align*} 

*Proof:*
We first verify this for $f = \chi_E$ where $E \in \mathcal M$.


We have $\tau_h f(x) = f(x-h) = \chi_E(x-h) = \chi_{E + h}(x)$ and 
$$
\int \tau_n f = m(E+h) = m(E) = \int f
,$$ 
where we know the measures are equal by translation invariance of measure.

By linearity, this holds for simple functions as well.

> Useful technique: once you know something for simple functions, you can often apply MCT to get it for $L^+$ functions as well!

If now $f\in L^+$ then there exists a sequence of simple functions $\theset{\phi_k} \nearrow f$, and by the MCT, 
$$
\int \phi_k \to \int f
.$$

Note that 
$$
\theset{\tau_h \phi_k} \nearrow \tau_h f
,$$ so 
$$
\int \tau_h \phi_k \to \int \tau_h f
.$$

But $\theset{\int\tau_h \phi_k} = \theset{\int\phi_k}$, so **by uniqueness of limits** we must have 
$$
\int f = \int \tau_h f
.$$

Now this follow for $\RR\dash$valued functions by writing $f = f_+ - f_-$, and then for $\CC\dash$valued functions by $f = \Re(f) + i \Im(f)$.

$\qed$

## Agreement of Riemann and Lebesgue Integrals

**Theorem:**
Let $f$ be a bounded $\RR\dash$valued function on a closed interval $[a,b]$.

If $f$ is Riemann integrable, then $f \in L^1$ (so $\mathcal{R} \subseteq L^1$ is a subspace) and the integrals agree, so
$$
\int_a^b f(x)~dx = \int_{[a,b]} f(x)~dx
$$

*Proof:*
Given a partition $P = \theset{t_1, t_2, \cdots t_n}$ of $[a,b]$, let 

\begin{align*}
G_p &= \sum_{j=1}^n \sup \theset{f(x) \suchthat x \in [t_k, t_{j+1}]} \chi_{[t_j, t_{j+1}]} \\
g_p &= \sum_{j=1}^n \inf \theset{f(x) \suchthat x \in [t_k, t_{j+1}]} \chi_{[t_j, t_{j+1}]}
.\end{align*}


Then $\int G_p = U(f, P)$ and $\int g_p = L(f, p)$ where $U, L$ denote the upper and lower sums.
Note that the Riemann integral is the infimum of the former and the supremum of the latter, over increasingly fine partitions.

So let $\theset{P_k}$ be a sequence of partitions with the size of the mesh going to 0.

Then $G_{P_k} \searrow G$ is converging to something, and $g_{p_k} \nearrow g$.
In particular, we have 
$$
G_P \leq f \leq g_P \quad \text{ and so }\quad G \leq f \leq g
.$$

Since $f$ is bounded, say by $M$, then both of these sequences are dominated by $\pm M \chi_{[a, b]} \in L^1$. 

So we can invoke the DCT, which yields 
$$
\int G_{P_k} \to \int G \text{ and } \int g_{P_k} \to \int g
,$$ 
and thus 
$$
\int g = \int G = \int_\mathcal{R} f
.$$

Since 
$$
\int G = \int g \implies \int (G-g) = 0 \implies G = g \text{ a.e. }
,$$
we have $f = G$ a.e.

But $G$ is a sequence of measurable function, and so $f$ is measurable.
Moreover, $\int G = \int f$. But $\int G = \int_\mathcal{R} f$ as well, so the two integrals agree.

> Recall that 
$$
f = 0 \text{ a.e. } \iff \int_E f = 0 \text{ for all } E \subseteq \mathcal{M}
.$$

$\qed$

Examples next time: 

Continuous functions with compact support are dense in $L^1$, which is a version of the following:

**Littlewood's  Principle**: Any integrable function is *almost * continuous, in the sense that for any $f$ and any $\varepsilon >0$ there is a continuous function $g$ such that 
$$
\int f - \int g < \varepsilon
.$$