Measures | Mathematics - Wyatt's Notes

A measure on a measurable space $(X, \mathcal{F})$ is a function $\mu : \mathcal{F} \to [0, \infty]$ satisfying:

1. $\mu(\varnothing) = 0$.
2. Countable additivity: if $\{A_n\}_{n=1}^{\infty}$ are pairwise disjoint sets in $\mathcal{F}$, then

$\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \sum_{n=1}^{\infty} \mu(A_n)$

The triple $(X, \mathcal{F}, \mu)$ is called a measure space.

Proposition 2.1 (Monotonicity). If $A \subseteq B$, then $\mu(A) \leq \mu(B)$.

Proof. $B = A \cup (B \setminus A)$ is a disjoint union, so $\mu(B) = \mu(A) + \mu(B \setminus A) \geq \mu(A)$. $\blacksquare$

Proposition 2.2 (Countable Subadditivity). For any sequence $\{A_n\} \subseteq \mathcal{F}$:

$\mu\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} \mu(A_n)$

Proof. Define $B_1 = A_1$ and $B_n = A_n \setminus \bigcup_{k=1}^{n-1} A_k$ for $n \geq 2$. Then $\{B_n\}$ are pairwise disjoint with $\bigcup B_n = \bigcup A_n$. By countable additivity and monotonicity, $\mu(\bigcup A_n) = \sum \mu(B_n) \leq \sum \mu(A_n)$. $\blacksquare$

Proposition 2.3 (Continuity from Below). If $A_1 \subseteq A_2 \subseteq \cdots$, then

$\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)$

Proof. Write $\bigcup A_n = A_1 \cup (A_2 \setminus A_1) \cup (A_3 \setminus A_2) \cup \cdots$, a disjoint union. Then $\mu(\bigcup A_n) = \mu(A_1) + \sum_{n=1}^{\infty} \mu(A_{n+1} \setminus A_n) = \lim_{n \to \infty} \mu(A_n)$. $\blacksquare$

Proposition 2.4 (Continuity from Above). If $A_1 \supseteq A_2 \supseteq \cdots$ and $\mu(A_1) < \infty$, then

$\mu\left(\bigcap_{n=1}^{\infty} A_n\right) = \lim_{n \to \infty} \mu(A_n)$

Example (Counting Measure). On any set $X$ with $\mathcal{F} = \mathcal{P}(X)$, define $\mu(A) = |A|$ (the cardinality, $\infty$ for infinite sets). This is a measure.

Example (Dirac Measure). For $x_0 \in X$, define $\delta_{x_0}(A) = 1$ if $x_0 \in A$, and $0$ otherwise. This is a measure.

Example (Lebesgue Measure). The Lebesgue measure $m$ on $\mathbb{R}$ is the completion of a measure on $\mathcal{B}(\mathbb{R})$ satisfying $m([a, b]) = b - a$ for all $a \leq b$. On $\mathbb{R}^n$, the Lebesgue measure satisfies $m([a_1, b_1] \times \cdots \times [a_n, b_n]) = \prod_{i=1}^n (b_i - a_i)$.

Example (Hausdorff Measure). For $\alpha > 0$, the $\alpha$-dimensional Hausdorff measure $\mathcal{H}^\alpha$ on $\mathbb{R}^n$ generalises Lebesgue measure to non-integer dimensions: $\mathcal{H}^\alpha(A) = \lim_{\delta \to 0} \inf\{\sum_i (\operatorname{diam} U_i)^\alpha : A \subseteq \bigcup U_i, \operatorname{diam} U_i < \delta\}$. For $\alpha = n$, $\mathcal{H}^n$ coincides with the Lebesgue measure up to a constant factor.

Definition. A set $N \subseteq X$ is null (or $\mu$-negligible) if $\mu(N) = 0$. A property holds almost everywhere (a.e.) if it holds on the complement of a null set.

Proposition 2.5. A countable union of null sets is null.

Definition. A measure space $(X, \mathcal{F}, \mu)$ is complete if every subset of a null set is measurable (and hence null). The Lebesgue measure is the completion of the Borel measure.

Theorem 2.6 (Completion). Every measure $\mu$ on a $\sigma$-algebra $\mathcal{F}$ has a unique completion $\overline{\mu}$ on the $\sigma$-algebra $\overline{\mathcal{F}} = \{A \cup N : A \in \mathcal{F}, N \subseteq B \in \mathcal{F}, \mu(B) = 0\}$.

Theorem 2.7. Given $\sigma$-finite measure spaces $(X, \mathcal{F}, \mu)$ and $(Y, \mathcal{G}, \nu)$, there exists a unique product measure $\mu \times \nu$ on $(X \times Y, \mathcal{F} \otimes \mathcal{G})$ satisfying $(\mu \times \nu)(A \times B) = \mu(A)\nu(B)$ for all $A \in \mathcal{F}$, $B \in \mathcal{G}$.

Theorem 2.8 (Fubini-Tonelli). If $f(x, y)$ is nonnegative and measurable (Tonelli) or integrable (Fubini), then:

$\int_{X \times Y} f\, d(\mu \times \nu) = \int_X \left(\int_Y f(x, y)\, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x, y)\, d\mu(x)\right) d\nu(y)$

Definition. A signed measure $\nu$ on $(X, \mathcal{F})$ is a countably additive function $\nu : \mathcal{F} \to (-\infty, \infty]$ that can take at most one of the values $\pm\infty$.

Theorem 2.9 (Hahn Decomposition). For any signed measure $\nu$, there exists a Hahn decomposition $X = P \cup N$ where $P$ is positive (every measurable subset has $\nu \geq 0$) and $N$ is negative (every subset has $\nu \leq 0$). This decomposition is unique up to null sets.

Theorem 2.10 (Jordan Decomposition). Every signed measure $\nu$ can be uniquely expressed as $\nu = \nu^+ - \nu^-$ where $\nu^+$ and $\nu^-$ are positive measures called the positive and negative variations.

Problem. Show that the Vitali set is not Lebesgue measurable.

Problem. Compute the Lebesgue measure of the Cantor set $C$.

flowchart TD
    A[2_Measures] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

A measure assigns a “size” to sets in a way that is consistent under subdivision. The three axioms — non-negativity, the measure of the empty set is zero, and countable additivity for disjoint sets — capture the idea that the whole equals the sum of its parts, even when there are infinitely many. Lebesgue measure on the real line generalises length: intervals get their usual length, and more complicated sets are measured by covering them with intervals. The Cantor set illustrates the subtlety: it is uncountable yet has measure zero, showing that cardinality and measure are independent notions. The $\sigma$-algebra determines which sets are measurable — a technical but essential requirement for consistency.

Mistake 1: Confusing measures with probability measures A probability measure satisfies $\mu(X) = 1$, but a general measure can take any value in $[0, \infty]$. Not every measure space is a probability space. For example, Lebesgue measure on $\mathbb{R}$ satisfies $m(\mathbb{R}) = \infty$. Always check whether the problem requires a probability measure or a general measure.

Mistake 2: Assuming countable additivity implies finite additivity Countable additivity is strictly stronger than finite additivity. A finitely additive set function need not be countably additive. For example, the “density” set function on $\mathbb{N}$ (proportion of elements in a set) is finitely additive but not countably additive. Measure theory requires countable additivity as a fundamental axiom.

Mistake 3: Forgetting that $\sigma$-algebras must be closed under complements and countable unions A collection of sets closed under finite unions and complements is an algebra, not necessarily a $\sigma$-algebra. The Borel $\sigma$-algebra is generated by open sets and is closed under countable operations. Many constructions in measure theory fail if you only have an algebra rather than a $\sigma$-algebra.

• Sigma-Algebras and Measurable Spaces: Measures are defined on sigma-algebras, which determine which subsets of a space can be measured.

• Lebesgue Integration: The Lebesgue integral is defined using measures and is essential for the convergence theorems of integration theory.

• Riemann Integration: Lebesgue’s criterion characterises Riemann integrability in terms of the measure of the discontinuity set.

• Quantum Mechanics

• Graph Theory

• Classical Mechanics

• Electromagnetism

• Statistical Learning

• Statistical Mechanics