## Tuesday Lecture

**Definition:**
A simple function $\RR^n \to \RR$ is a finite $\CC\dash$linear combination of characteristic functions of measurable sets, i.e.
$$
s(x) = \sum_{j=1}^N a_j \chi_{E_j}.
$$

Equivalently, $f: \RR^n \to \CC$ is simple $\iff$ $f$ is measurable and the range of $f$ is a *finite* subset of $\CC$.

This lets us write $E_j = f\inv(\theset a_j)$ are all disjoint and the $a_j$ are distinct, and it's not a big jump to also require $\union E_j = \RR^n$ by including the set where $f$ is zero.
This is *standard representation* of a simple function $f$.

**Theorem (Approximation by Simple Functions):**
If $f: E \to [0, \infty]$ is measurable, then there exists a sequence of simple functions $\theset{\phi_i}_{i=1}^\infty$ such that

- $\phi_k(x) \leq \phi_{k+1}(x) \forall x, \forall k$, and

- $\phi_k(x) \to f(x)$ pointwise $\forall x$.

*Proof:*
First take an initial approximation.
For each $k\in \NN$, let $Q_k \definedas [-\frac k 2, \frac k 2]^n$.
Now let

\begin{align*}
F_k(x) = \begin{cases}
f(x) & x\in Q_k \text{ and } f(x) \leq k, \\
k & x\in Q_k \text{ and } f(x) > k, \\
0 & \text{otherwise}.
\end{cases}
\end{align*}

Then $\lim_{k\to\infty} F_k(x) = f(x)$ for all $x$.

Now break the vertical distance $k$ up into pieces of width $\frac 1 k$, and for each $0 \leq j \leq k^2$, define
$$
E_{j, k} \definedas \theset{x \in Q_k \suchthat \frac j k < F_k (x) \leq \frac{j+1} k}.
$$

![Image](figures/2019-09-17-11:25.png)\

Now let
$$
\phi_k(x) = \sum_{j=0}^{k^2} \frac j k \chi_{E_{k, j}}.
$$

Then $\phi_k(x)$ is a simple function that is increasing to $f$, and we have
$$
\abs{f(x) - \phi_k(x)} \leq \abs{f(x) - F_k(x)} + \abs{F_k(x) - \phi_k(x)} \leq 0 + \frac 1 k \to 0.
$$

## Lebesgue Integration


### Step 1: Simple Functions

**Step 1:** 
Define $\int f$ for $f$ non-negative and simple.

Let 
$$
L^+ \definedas \theset{f: \RR^n \to [0, \infty] \suchthat f \text{ is measurable }}.
$$

If $\phi \in L^+$ is written in the standard representation, i.e. $\phi(x) = \sum_{i=1}^N a_i \chi_{E_i}$, then define

\begin{align*}
\int \phi \definedas \int \phi(x) ~dx = \sum_{j=1}^N a_j m(E_j)
.\end{align*}


where we may have $a_j = 0$ and $m(E_j) = \infty$ for some $j$, so we do this with the understanding that $0 \times \infty = 0$.

Note that if $A$ is measurable, then $\phi \chi_A$ is also simple and 
$$
\int_A \phi = \int \phi \chi_A.
$$

### Properties of the Lebesgue Integral

1. If $\phi, \psi \in L^+$ are simple, then
    a. If $c \geq 0$ then $\int c\phi$ is well-defined and equal to $c\int \phi$. 
    
        *(Easy to show)*

2. $\int (\psi + \phi) = \int \psi + \int \phi$.

3. If $\psi \leq \phi$, then $\int \psi \leq \int \phi$.

4. The map $A \mapsvia{\mu_A} \int_A \phi$ is a *measure* 

    So $\phi$ acts like a *density*.

*Proof of (1):* 
Easy exercise.

*Proof of (2):*
Write $\phi, \psi$ in standard representation, so

\begin{align*}
\phi = \sum a_i \chi_{E_i} \\
\psi = \sum b_i \chi_{F_i}
.\end{align*}


Then note that 
$$
E_j = \union_{k=1}^M E_i \intersect F_k
,$$ 
and each of these terms are disjoint since the $F_k$ were disjoint, and so

\begin{align*}
\int \phi = \sum_{j=1}^n a_j m(E_j) = \sum_{j=1}^n \sum_{i=1}^M a_j m(E_j \intersect F_k) \\
\int \psi = \sum_{j=1}^n b_j m(F_j) = \sum_{j=1}^n \sum_{i=1}^M b_j m(E_j \intersect F_k)
.\end{align*}


But note that while $\phi = \sum_{j, k} a_j \chi_{E_j \intersect F_k}$ is not a standard representation, and likewise for $\psi$, but the integral takes the expected form. 

So we have
$$
\int\psi + \int\phi = \sum_{j, k}(a_j + b_j)m(E_j \intersect F_k).
$$

But consider the LHS of the original equation. 
We have 
$$
\phi + \psi = \sum_{j, k} (a_j + b_k) \chi_{E_j \intersect F_k}
,$$ 
which is almost in standard form. 
The sets are disjoint, but it may occur that $a_j + b_k$ are not distinct. 

But if (for example) the value 2 occurs twice in these values, and there are two sets in the preimage, we can just take their union. 

Since the sets are disjoint, the measure of the union is the sum of the measures, and so this quantity is still $\int (\psi + \phi)$ as desired.
$\qed$

*Proof of (3):*
If $\phi \leq \psi$, then $a_j \leq b_k$ when $E_j \intersect F_k \neq \emptyset$.

Then,
$$
\int \phi = \sum_{j, k} a_i m(E_j \intersect F_k) \leq \sum_{j, k} b_k m(E_j \intersect F_k) = \int \psi.
$$

*Proof of (4):*
Suppose $A\in \mathcal{M}$, so $A$ is measurable.

We want to show that
$$
\mu_A \definedas \int_A \phi
,$$ 
is countably additive.

So suppose $A = \disjoint A_k$ is a disjoint union of measurable sets.
We need to show 
$$
\int_A \phi = \sum_k \int_{A_k}\phi
.$$

We have

\begin{align*}
\int_A \phi = \sum a_i m(E_i \intersect A) = \sum_k \sum_j a_j m(E_j \intersect A_k) = \sum_k \int_{A_k} \phi
.\end{align*}


where we note that $E_j \intersect A = \union_k E_j \intersect A_k$, which is a disjoint union of measurable sets.

$\qed$

### Step 2: Non-Negative Functions

Extending the definition to all of $L^+$:

If $f\in L^+$, we define
$$
\int f = \sup\theset{\int \phi \suchthat 0 \leq \phi \leq f,~ \phi \text{ simple }}.
$$

Note that if $f$ itself is simple, then by property (3), this definition agrees with the previous definition.

We also have

1. $f \leq g \implies \int f \leq \int g$, since we're taking the supremum over a larger set on the RHS.

2. $c\geq 0 \implies \int cf = c\int f$.

> Note that it is not obvious that $\int (f+g) = \int f + \int g$ for $f,g \in L^+$.

#### Monotone Convergence Theorem

**Incredibly Important Theorem (Monotone Convergence):**
If $\theset f_n \subset L^+$ with $f_n \leq f_{n+1}$ for all $n$ and $f_n \to f$ pointwise, then
$$
\lim \int f_n = \int \lim f_n \coloneqq \int f.
$$

**Consequence:** 
If $f,g \in L^+$, then $\exists \phi_n \nearrow f, \psi_n \nearrow g$, and so $\phi_n + \psi_n \nearrow f + g$ and

\begin{align*}
\int(f + g) &=_{\text{MCT}} \lim \int (\phi_n + \psi_n) \\
&= \lim( \int \phi_n + \int \psi_n ) \quad\quad\text{ since $\phi_n,\psi_n$ are simple } \\
&= \lim \int \phi_n + \lim \int \psi_n \\
&=_{\text{MCT}} \int f + \int g.
.\end{align*}