# Math4121 Lecture 32

## Chapter 6: The Lebesgue Integral

### Measurable Functions

Definition: A function $f:\mathbb{R}\to\mathbb{R}$ is measurable on the interval $[a,b]$ if $\{x\in [a,b]:f(x) > c\}$ is measurable for all $c\in\mathbb{R}$, called the **super level set** of $f$.

Denote $\{f> c\}$

#### Proposition 6.1

The following are equivalent:

For all $c\in\mathbb{R}$,

1. $\{x\in [a,b]:f(x) > c\}$ is measurable.
2. $\{x\in [a,b]:f(x) < c\}$ is measurable.
3. $\{x\in [a,b]:f(x) \geq c\}$ is measurable.
4. $\{x\in [a,b]:f(x) \leq c\}$ is measurable.
5. $\{x\in \mathbb{R}:c \leq f(x) < d\}$ is measurable for all $c,d\in\mathbb{R}$.

Proof:

Since the complement of a measurable set is measurable. (1) $\iff$ (4). and (2) $\iff$ (3).

We only need to show (1) $\implies$ (2).

Since $\{f>c\}=\bigcup_{n=1}^{\infty}\{f\geq c+\frac{1}{n}\}$.

So (1) $\implies$ (2).

Since (2) $\implies$ (1)-(4) $\implies$ (5).

To see (5) $\implies$ (1), we have $\{f\geq c\}=\bigcup_{n=1}^{\infty}\{x\in\mathbb{R}:c \leq f(x) < c+n\}$

QED

#### Proposition 6.3

Let $f,g$ be measurable on $[a,b]$ and $\alpha\in\mathbb{R}$. Then the following are measurable:

1. $f+g$
2. $fg$
3. $\alpha f$
4. $|f|^\alpha$

Proof:

If $\alpha=0$, then $\alpha f$ and $|f|^\alpha$ are constant functions, hence measurable.

But for constant functions $h$, $\{h>c\}=\begin{cases}
\emptyset & \text{if } c\geq h \\
\mathbb{R} & \text{if } c < h
\end{cases}$

For $a\neq 0$, $\{x\in \mathbb{R}:\alpha f(x) > c\}=\{x\in \mathbb{R}:f(x) > \frac{c}{\alpha}\}$

Similarly, $\{x\in \mathbb{R}:|f(x)|^\alpha > c\}=\{x\in \mathbb{R}:|f(x)| > c^{1/\alpha}\}=\{x\in \mathbb{R}:f(x) > c^{1/\alpha}\}\cup\{x\in \mathbb{R}:f(x) < -c^{1/\alpha}\}$

We want to show $\{f+g>c\}=\bigcup_{q\in \mathbb{Q}}\{f>q\}\cap\{g>c-q\}$

$\{f+g>c\}\supseteq \bigcup_{q\in \mathbb{Q}}\{f>q\}\cap\{g>c-q\}$

if $x$ is in the RHS, then $\exists q\in \mathbb{Q}$ such that $f(x) > q$ and $g(x) > c-q$, therefore $f(x)+g(x) > c$.

So $x$ is in the LHS.

$\{f+g>c\}\subseteq \bigcup_{q\in \mathbb{Q}}\{f>q\}\cap\{g>c-q\}$

Let $x\in \mathbb{R}$ such that $f(x)+g(x) > c$. Need to find $q\in \mathbb{Q}$ such that $f(x) > q$ and $g(x) > c-q$.

Since $f(x)>c-g(x)$, by the density of $\mathbb{Q}$ in $\mathbb{R}$, $\exists q\in \mathbb{Q}$ such that $q > c-g(x)$.

For $fg$, we have $fg=\frac{1}{4}((f+g)^2-(f-g)^2)$

So $fg$ is measurable.

QED

### Limit of Measurable Functions

#### Proposition 6.4

Let $\{f_n\}$ be a sequence of measurable functions on $[a,b]$. Then,

$$
g(x)=\sup_{n\geq 1}f_n(x),\inf_{n\geq 1}f_n(x),\limsup_{n\to\infty}f_n(x),\liminf_{n\to\infty}f_n(x)
$$

are all measurable functions

#### Corollary of Proposition 6.4

If $\{f_n\}_{n=1}^{\infty}$ are measurable functions and $f(x)=\lim_{n\to\infty}f_n(x)$ exists for all $x\in[a,b]$, then $f$ is measurable. (pointwise limit of measurable functions is measurable)

#### Definition of almost everywhere

A property holds **almost everywhere** if it holds everywhere except for a set of measure zero.