Analysis

Measure theory

A rigorous notion of size for sets, underlying the Lebesgue integral and modern probability.

IntuitionWhy 'length' needs a rigorous definition

The length of an interval $[a,b]$ is obviously $b-a$ , but what is the 'length' (or area, or volume) of a wild set like the rationals in $[0,1]$ , or a fractal? Measure theory answers this by first choosing a family of 'well-behaved' subsets — called a $\sigma$ -algebra — on which a consistent notion of size, a measure $\mu$ , can be defined; famously, no measure can be defined consistently on the family of all subsets of $\mathbb{R}$ (as the Vitali set construction shows), so restricting to a suitable $\sigma$ -algebra is unavoidable, not a technicality.

Refining a Riemann sum's partition of the

x

-axis to

n=55

subintervals: for smooth functions the sum converges nicely, but for wildly discontinuous functions (like the indicator of the rationals) no amount of refinement helps — Lebesgue's fix is to partition the

y

-axis (the range) into level sets instead, and measure those.

UndergraduateFormal definitions: $\sigma$ -algebras, measures, and the Lebesgue integral

Definition: $\sigma$ -algebra and measure space

A $\sigma$ -algebra $\mathcal{F}$ on a set $X$ is a family of subsets containing $X$ itself, closed under complements, and closed under countable unions. A measure $\mu : \mathcal{F} \to [0,\infty]$ assigns $\mu(\emptyset)=0$ and is countably additive: the measure of a countable disjoint union equals the sum of the individual measures. The triple $(X, \mathcal{F}, \mu)$ is called a measure space; when $X=\mathbb{R}^n$ , $\mathcal{F}$ is the family of Lebesgue-measurable sets, and $\mu$ assigns $[a,b]$ the value $b-a$ , we call $\mu$ Lebesgue measure.

\mu\left(\bigcup_{n=1}^{\infty} A_n\right) = \sum_{n=1}^{\infty} \mu(A_n) \quad \text{for pairwise disjoint } A_1, A_2, \ldots \in \mathcal{F}

A function $f : X \to \mathbb{R}$ is measurable if $f^{-1}((a,\infty)) \in \mathcal{F}$ for every real $a$ . The Lebesgue integral is built up in stages: first for simple functions $s = \sum_{k} c_k \mathbf{1}_{E_k}$ (finite linear combinations of indicators of measurable sets), where $\int s\,d\mu := \sum_k c_k \mu(E_k)$ ; then for general nonnegative measurable $f$ , by approximating from below with simple functions.

\int f\,d\mu = \sup\left\{ \int s\,d\mu : 0 \le s \le f,\ s \text{ simple and measurable} \right\}

Riemann integral versus Lebesgue integral
Aspect	Riemann integral	Lebesgue integral
Partition strategy	Partitions the domain (the $x$ -axis) into subintervals	Partitions the range (the $y$ -axis) into level sets
Class of integrable functions	Bounded functions continuous almost everywhere	Any measurable function (e.g. the Dirichlet function)
Limit-interchange theorems	Requires uniform convergence in general	Monotone and Dominated Convergence Theorems suffice

UndergraduateCore theorems: Monotone Convergence and Dominated Convergence

Theorem: The Monotone Convergence Theorem

Let $(f_n)$ be measurable functions with $0 \le f_1 \le f_2 \le \cdots$ pointwise almost everywhere, and let $f = \lim_{n\to\infty} f_n$ pointwise almost everywhere. Then $\lim_{n\to\infty} \int f_n\,d\mu = \int f\,d\mu$ .

Why is it true?

Since the $f_n$ only grow, no 'area' is ever lost or double-counted as $n$ increases; stacking their simple-function approximations, the total accumulated area converges exactly to the area under the limit $f$ .

Proof

Since $\int f_n\,d\mu$ is a nondecreasing sequence of extended reals (as $f_n$ increases), the limit $L := \lim_{n\to\infty}\int f_n\,d\mu$ exists in $[0,\infty]$ . Since $f_n \le f$ for every $n$ , monotonicity of the integral gives $\int f_n\,d\mu \le \int f\,d\mu$ for every $n$ , hence $L \le \int f\,d\mu$ .

For the reverse inequality, fix any simple measurable function $s$ with $0 \le s \le f$ and fix a constant $c \in (0,1)$ . Define $E_n = \{x : f_n(x) \ge c\,s(x)\}$ . Because $f_n$ increases to $f \ge s$ pointwise, the sets $E_n$ increase and their union is (almost) all of $X$ : for a.e. $x$ with $s(x)=0$ this is automatic, and for a.e. $x$ with $s(x)>0$ , eventually $f_n(x) > c\,s(x)$ since $f_n(x)\to f(x) \ge s(x) > c\,s(x)$ .

Writing $s = \sum_k c_k \mathbf{1}_{E_k}$ , we get $\int f_n\,d\mu \ge \int_{E_n} f_n\,d\mu \ge c\int_{E_n} s\,d\mu = c\sum_k c_k\,\mu(E_k \cap E_n)$ . By countable additivity (continuity of measure from below, since $E_n$ increases to $X$ ), $\mu(E_k \cap E_n) \to \mu(E_k)$ as $n\to\infty$ , so letting $n\to\infty$ gives $L \ge c\sum_k c_k\,\mu(E_k) = c\int s\,d\mu$ .

Letting $c \to 1$ gives $L \ge \int s\,d\mu$ for every simple $s \le f$ , and taking the supremum over all such $s$ (the very definition of $\int f\,d\mu$ ) gives $L \ge \int f\,d\mu$ . Combined with $L \le \int f\,d\mu$ from above, we conclude $L = \int f\,d\mu$ .

Theorem: The Dominated Convergence Theorem

Let $(f_n)$ be measurable functions with $f_n \to f$ pointwise almost everywhere, and suppose there is an integrable function $g$ (i.e. $\int g\,d\mu < \infty$ ) with $|f_n| \le g$ almost everywhere for every $n$ . Then $f$ is integrable and $\lim_{n\to\infty} \int f_n\,d\mu = \int f\,d\mu$ .

Why is it true?

The dominating function $g$ acts as a fixed 'ceiling' of finite total mass that no $f_n$ can exceed, ruling out mass silently escaping to infinity — the exact failure mode that defeats naive limit-interchange arguments when there is no monotonicity to rely on.

Proof

Since $g - f_n \ge 0$ almost everywhere, Fatou's lemma applies: $\int \liminf_{n\to\infty}(g-f_n)\,d\mu \le \liminf_{n\to\infty} \int (g-f_n)\,d\mu$ . Because $f_n \to f$ , the left side equals $\int (g-f)\,d\mu = \int g\,d\mu - \int f\,d\mu$ (valid since $\int g\,d\mu$ is finite), and the right side equals $\int g\,d\mu - \limsup_{n\to\infty}\int f_n\,d\mu$ .

Cancelling the finite quantity $\int g\,d\mu$ from both sides gives $-\int f\,d\mu \le -\limsup_{n\to\infty}\int f_n\,d\mu$ , i.e. $\limsup_{n\to\infty}\int f_n\,d\mu \le \int f\,d\mu$ .

Next, since $g+f_n \ge 0$ almost everywhere, Fatou's lemma applied to $(g+f_n)$ gives $\int (g+f)\,d\mu \le \liminf_{n\to\infty}\int (g+f_n)\,d\mu$ , and cancelling $\int g\,d\mu$ again gives $\int f\,d\mu \le \liminf_{n\to\infty}\int f_n\,d\mu$ .

Combining the two inequalities, $\limsup_{n\to\infty}\int f_n\,d\mu \le \int f\,d\mu \le \liminf_{n\to\infty}\int f_n\,d\mu$ . Since $\liminf \le \limsup$ always holds, all three quantities must coincide, so $\lim_{n\to\infty}\int f_n\,d\mu$ exists and equals $\int f\,d\mu$ .

UndergraduateReal-World Applications and Worked Examples

Measure theory is the invisible foundation of modern probability: Kolmogorov's axioms (1933) define a random experiment as a measure space of total mass $1$ , so that expectation is literally a Lebesgue integral; in signal processing and Fourier analysis, the $L^p$ spaces of measurable functions (built from the Lebesgue integral) are the natural setting for Parseval's identity and Fourier series convergence; in mathematical finance, stochastic integration and option pricing rely on measure-theoretic probability; and fractal geometry and image compression use generalizations of Lebesgue measure (Hausdorff measure) to assign a meaningful 'size' to sets with non-integer dimension.

Example: The Dirichlet function: Lebesgue integrable, not Riemann integrable

Let $f = \mathbf{1}_{\mathbb{Q} \cap [0,1]}$ be the indicator of the rationals in $[0,1]$ (the Dirichlet function). Show $f$ is not Riemann integrable, but is Lebesgue integrable with $\int_0^1 f\,d\mu = 0$ .

Solution

For Riemann integrability: on any subinterval of any partition of $[0,1]$ , both rationals and irrationals are dense, so the supremum of $f$ on that subinterval is $1$ and the infimum is $0$ . Hence every upper Riemann sum equals $1$ and every lower Riemann sum equals $0$ , regardless of how fine the partition is; since these never converge to a common value, $f$ is not Riemann integrable.

For Lebesgue integrability: the set $\mathbb{Q}\cap[0,1]$ is countable, so enumerate it as $\{q_1,q_2,\ldots\}$ . By countable subadditivity of Lebesgue measure, $\mu(\mathbb{Q}\cap[0,1]) \le \sum_{i=1}^\infty \mu(\{q_i\}) = \sum_{i=1}^\infty 0 = 0$ , so $\mathbb{Q}\cap[0,1]$ has Lebesgue measure zero.

Since $f$ equals the zero function everywhere except on this measure-zero set, and the Lebesgue integral is unaffected by modifying a function on a set of measure zero, $\int_0^1 f\,d\mu = \int_0^1 0\,d\mu = 0$ .

Example: Evaluating a limit of integrals with the Monotone Convergence Theorem

Let $f_n(x) = \left(1-\frac{x}{n}\right)^n$ for $0 \le x \le n$ and $f_n(x) = 0$ for $x > n$ . Using the Monotone Convergence Theorem, evaluate $\lim_{n\to\infty} \int_0^\infty f_n\,d\mu$ .

Solution

For each fixed $x \ge 0$ , it is a standard calculus fact that $\left(1-\frac{x}{n}\right)^n \to e^{-x}$ as $n\to\infty$ (take logarithms and use $n\ln(1-x/n) \to -x$ ), and that $f_n(x)$ is nondecreasing in $n$ for fixed $x$ (a consequence of the concavity of $\ln$ ). Thus $0 \le f_1 \le f_2 \le \cdots$ and $f_n \to e^{-x}\mathbf{1}_{[0,\infty)}(x)$ pointwise.

By the Monotone Convergence Theorem, $\lim_{n\to\infty} \int_0^\infty f_n\,d\mu = \int_0^\infty e^{-x}\,dx$ .

This is a standard improper Riemann integral (which agrees with the Lebesgue integral here, since $e^{-x} \ge 0$ ): $\int_0^\infty e^{-x}\,dx = \left[-e^{-x}\right]_0^\infty = 0 - (-1) = 1$ . So $\lim_{n\to\infty} \int_0^\infty f_n\,d\mu = 1$ .

Research frontier as of 2026

The classical Lebesgue Differentiation Theorem guarantees that, for an integrable $f$ , averaging $f$ over shrinking balls around almost every point recovers $f$ 's value there. Whether the analogous statement holds when averaging over long thin tubes pointing in every possible direction is governed by the size of Kakeya sets: sets of Lebesgue measure zero that nonetheless contain a full unit line segment in every direction, first constructed in 1920 by Abram Besicovitch while studying exactly this kind of differentiation question. How small such sets can be in a dimension-theoretic sense — the Kakeya conjecture — stayed open in $\mathbb{R}^3$ for over a century, until Hong Wang and Joshua Zahl's 2025 preprint proved that every Kakeya set in $\mathbb{R}^3$ must have full Hausdorff dimension $3$ , a landmark result in geometric measure theory with consequences reaching into harmonic analysis and partial differential equations.

What is the Lebesgue integral $\int_0^1 f\,d\mu$ of the Dirichlet function $f = \mathbf{1}_{\mathbb{Q}\cap[0,1]}$ ?

By countable additivity, what is the Lebesgue measure of $(0,1) \cup (2,3)$ ?

In Kolmogorov's measure-theoretic axioms for probability, what does a random experiment correspond to?

Which theorem justifies $\lim_{n\to\infty}\int f_n\,d\mu = \int f\,d\mu$ when $f_n \to f$ pointwise and $|f_n| \le g$ for a single fixed integrable $g$ , without requiring the $f_n$ to be monotone?

References

Gerald B. Folland (1999). Real Analysis: Modern Techniques and Their Applications
Donald L. Cohn (2013). Measure Theory
Hong Wang, Joshua Zahl (2025). Volume estimates for unions of convex sets, and the Kakeya set conjecture in three dimensions · arXiv:2502.17655 [preprint, not peer-reviewed]