Measurable Functions

5.1 Definition

Let $(X, \mathcal{F})$ and $(Y, \mathcal{G})$ be measurable spaces. A function $f : X \to Y$ is measurable if $f^{-1}(G) \in \mathcal{F}$ for every $G \in \mathcal{G}$. When $Y = \mathbb{R}$, we equip $\mathbb{R}$ with $\mathcal{B}(\mathbb{R})$.

Proposition 5.1. $f : X \to \mathbb{R}$ is measurable if and only if $f^{-1}((a, \infty)) \in \mathcal{F}$ for every $a \in \mathbb{R}$.

Proof. Since $(a, \infty)$ generates $\mathcal{B}(\mathbb{R})$, the $\sigma$-algebra $f^{-1}(\mathcal{B}(\mathbb{R}))$ equals the $\sigma$-algebra generated by $f^{-1}((a, \infty))$. $\blacksquare$

Proposition 5.2. Compositions of measurable functions are measurable.

Proposition 5.3. If $f, g : X \to \mathbb{R}$ are measurable, then $f + g$, $fg$, $f/g$ (when defined), $|f|$, $\max(f, g)$, and $\min(f, g)$ are all measurable.

5.2 Simple Functions

A simple function is a finite linear combination of indicator functions:

$s(x) = \sum_{i=1}^{n} a_i \chi_{A_i}(x)$

where $a_i \in \mathbb{R}$ and $A_i \in \mathcal{F}$ are measurable sets.

Theorem 5.4 (Approximation Theorem). For every non-negative measurable function $f : X \to [0, \infty]$, there exists an increasing sequence of simple functions $0 \leq s_1 \leq s_2 \leq \cdots$ converging pointwise to $f$.

Proof. For each $n$, partition $[0, n)$ into $n \cdot 2^n$ subintervals of length $2^{-n}$. Define

$s_n(x) = \begin{cases} \frac{k-1}{2^n} & \text{if } \frac{k-1}{2^n} \leq f(x) < \frac{k}{2^n},\ k = 1, \ldots, n2^n \\ n & \text{if } f(x) \geq n \end{cases}$

Each $s_n$ is a simple function, $s_n \leq s_{n+1}$, and $s_n(x) \to f(x)$ for every $x$. $\blacksquare$

5.3 Egorov”s Theorem and Lusin’s Theorem

Theorem 5.5 (Egorov’s Theorem). Let $(X, \mathcal{F}, \mu)$ be a finite measure space and let $f_n : X \to \mathbb{R}$ be measurable functions converging pointwise to $f$ a.e. Then for every $\varepsilon > 0$, there exists $A \in \mathcal{F}$ with $\mu(A) < \varepsilon$ such that $f_n \to f$ uniformly on $X \setminus A$.

Theorem 5.6 (Lusin’s Theorem). Let $f : [a, b] \to \mathbb{R}$ be Lebesgue measurable. Then for every $\varepsilon > 0$, there exists a compact set $K \subseteq [a, b]$ with $m([a, b] \setminus K) < \varepsilon$ such that $f|_K$ is continuous.