Measurable Functions | Mathematics

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.

• Lebesgue Measurable Sets and Non-Measurable Sets: Establishes the measurable sets upon which the definition of measurable functions depends.

• $L^p$ Spaces: Defines function spaces using measurable functions with finite $p$-norms, directly building on the convergence modes studied here.

• Fubini and Tonelli Theorems: Applies measurability of functions on product spaces to justify interchanging the order of integration.

• Classical Mechanics

• Electromagnetism

flowchart TD
    A[5_Measurable Functions] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Measurable functions are the functions that play nicely with measure theory — they are built from simple functions by taking limits. Simple functions (finite linear combinations of indicator functions) are the “pixels” of measurable functions, approximating any function from below. The Borel $\sigma$-algebra is generated by open sets, ensuring that limits of measurable functions remain measurable. The layer cake representation decomposes a function into horizontal slices: the integral equals the integral of the measures of the superlevel sets. This viewpoint converts questions about functions into questions about sets, which is often easier. Convergence in measure is weaker than pointwise convergence but more robust for passing to limits.

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.

Mistake 1: Assuming convergence in measure implies pointwise convergence Convergence in measure only guarantees an a.e.\ convergent subsequence, not that the original sequence converges pointwise. The typewriter sequence (moving indicator functions) converges in measure to $0$ on $[0,1]$ but diverges pointwise everywhere. Students often conflate these two modes of convergence.

Mistake 2: Forgetting that Egorov’s theorem requires a finite measure space Egorov’s theorem states that pointwise a.e.\ convergence implies uniform convergence outside a set of small measure, but this requires $\mu(X) < \infty$. On $\mathbb{R}$ with Lebesgue measure, $f_n = \chi_{[n, n+1]}$ converges pointwise to $0$ but Egorov’s theorem fails because the measure space is infinite.

Mistake 3: Assuming that measurable functions are continuous or have nice properties Measurable functions can be highly discontinuous — in fact, a function is measurable if and only if it is a pointwise limit of simple functions. Students sometimes assume measurability implies continuity or boundedness, which is false. The indicator function of the rationals is measurable but discontinuous everywhere.

This section provides detailed coverage of advanced concepts, including full derivations, proofs, and extended examples.

Complete mathematical derivations and proofs are provided where appropriate. Each step is explained to ensure understanding of the underlying reasoning.

Advanced examples demonstrate the application of concepts to complex problems. These examples go beyond standard exam questions to develop deeper understanding.

This material connects to current research and advanced applications in the field. Understanding these connections provides context for the study material.

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