Measurable Functions

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.

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 $x$. $\blacksquare$

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.

Definition. A sequence of measurable functions $f_n$ converges in measure to $f$ if for every $\varepsilon > 0$:

$\lim_{n \to \infty} \mu(\{x : |f_n(x) - f(x)| \geq \varepsilon\}) = 0$

Theorem 5.7. If $f_n \to f$ a.e. on a finite measure space, then $f_n \to f$ in measure.

Proof. For any $\varepsilon > 0$, let $E_n = \bigcup_{k \geq n} \{|f_k - f| \geq \varepsilon\}$. Then $E_{n+1} \subseteq E_n$ and $\mu(\bigcap_n E_n) = 0$ by a.e. convergence. By continuity from above, $\mu(E_n) \to 0$, hence $\mu(\{|f_n - f| \geq \varepsilon\}) \to 0$. $\blacksquare$

The converse is false but there is a partial converse:

Theorem 5.8. If $f_n \to f$ in measure, then there exists a subsequence $f_{n_k}$ converging to $f$ a.e.

Definition. For $1 \leq p < \infty$, $f_n \to f$ in $L^p(\mu)$ if:

$\int |f_n - f|^p \, d\mu \to 0 \quad \text{as } n \to \infty$

Proposition 5.9. Convergence in $L^p$ implies convergence in measure.

Proof. By Chebyshev’s inequality: $\mu(\{|f_n - f| \geq \varepsilon\}) \leq \frac{1}{\varepsilon^p} \int |f_n - f|^p \, d\mu \to 0$. $\blacksquare$

Proposition 5.10. Convergence a.e. does not imply convergence in $L^p$, and vice versa.

Example. $f_n = n\chi_{(0, 1/n)}$ on $[0, 1]$ with Lebesgue measure. Then $f_n \to 0$ a.e. but $\int f_n = 1$, so $f_n \nrightarrow 0$ in $L^1$.

The relationships between convergence modes (on a finite measure space) are:

1. Uniform convergence $\implies$ pointwise convergence $\implies$ a.e. convergence.
2. A.e. convergence (on finite measure) $\implies$ convergence in measure.
3. $L^p$ convergence $\implies$ convergence in measure.
4. Convergence in measure $\implies$ existence of a.e. convergent subsequence.

Problem 1. Show that if $f$ is measurable and $g$ is continuous, then $g \circ f$ is measurable.

Solution. For any open set $U \subseteq \mathbb{R}$, $(g \circ f)^{-1}(U) = f^{-1}(g^{-1}(U))$. Since $g$ is continuous, $g^{-1}(U)$ is open, hence Borel. Since $f$ is measurable, the preimage is in $\mathcal{F}$. $\blacksquare$

Problem 2. Prove that the pointwise limit of measurable functions is measurable.

Solution. If $f_n \to f$ pointwise, then $\{x : f(x) > a\} = \bigcup_{m=1}^\infty \bigcap_{n=m}^\infty \{x : f_n(x) > a + 1/m\}$. Each inner set is measurable, so the countable union/intersection is measurable. $\blacksquare$

Problem 3. Construct an example of convergence in measure but not a.e.

Solution. Let $X = [0, 1]$ with Lebesgue measure. Arrange indicator functions of intervals $[0, 1], [0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3], [2/3, 1], \ldots$ For each $x$, $f_n(x) = 1$ infinitely often and $0$ infinitely often, so no pointwise convergence. But $\mu(\{f_n = 1\}) \to 0$, so convergence in measure holds. $\blacksquare$

Proposition 5.11. If $A_n \in \mathcal{F}$ and $A_n \uparrow A$ (i.e., $A_n \subseteq A_{n+1}$ and $\bigcup A_n = A$), then $\mu(A_n) \to \mu(A)$. This is continuity from below.

Proposition 5.12. If $A_n \in \mathcal{F}$ and $A_n \downarrow A$ with $\mu(A_1) < \infty$, then $\mu(A_n) \to \mu(A)$. This is continuity from above.

Theorem 5.13 (Layer Cake Representation). For a non-negative measurable function $f$:

$\int_X f\, d\mu = \int_0^\infty \mu(\{x : f(x) > t\})\, dt$

This formula is useful for computing integrals and for proving inequalities such as Chebyshev’s and the Marcinkiewicz interpolation theorem.

Problem 4. Prove the layer cake representation using Fubini’s theorem.

Problem 5. Show that if $f_n \to f$ in $L^p$, then $f_n \to f$ in measure, but the converse does not hold. Construct a counterexample.