Measure & Lebesgue Integration
σ-algebras, Lebesgue measure, and why the Lebesgue integral subsumes the Riemann one.
A σ-algebra $\mathcal A$ on $X$ is a collection of subsets closed under complements and countable unions, containing $X$. A measure $\mu: \mathcal A \to [0, \infty]$ is countably additive: $\mu(\bigsqcup_i A_i) = \sum_i \mu(A_i)$ for disjoint $A_i$.
Lebesgue measure on $\mathbb R$ extends "length of an interval" to a vastly larger σ-algebra, the Lebesgue measurable sets. (Not all subsets — assuming the axiom of choice, non-measurable sets exist, e.g., Vitali sets.)
The Lebesgue integral partitions the range of $f$ rather than the domain:
$$\int f\, d\mu = \sup\Big\{\int s\, d\mu : s \text{ simple, } 0 \leq s \leq f\Big\}.$$Powerful convergence theorems hold:
- Monotone convergence: $f_n \uparrow f$ (a.e.) $\Rightarrow \int f_n \to \int f$.
- Dominated convergence: $f_n \to f$ a.e. and $|f_n| \leq g$ with $\int g < \infty \Rightarrow \int f_n \to \int f$.
Lebesgue integrable functions form a complete normed space $L^1(\mu)$ — a property that fails for the Riemann theory.