Lebesgue Measure and the Lebesgue Integral

Why Was a New Integral Needed

By the end of the 19th century, mathematicians encountered the limitations of the Riemann integral. The first problem: functions with “bad” discontinuities. The Dirichlet function $D(x) = 1$ for rational $x$ and $D(x) = 0$ for irrational numbers is not Riemann integrable: for any partition of $[0,1]$ the lower Darboux sum is $0$ (there are irrational points in every interval), and the upper sum is $1$ (there are rational points in every interval). However, rational numbers are “rare”—their continuum cannot be detected by measurement, thus the “proper” integral of $D$ should be $0$.

The second problem: passage to the limit. If $f_n \to f$ and each $f_n$ is integrable, is $f$ integrable? Is it true that $\lim \int f_n = \int f$? For Riemann, the answer is “not necessarily”—uniform convergence is required. But in probability theory, functional analysis, and the theory of differential equations, we need much more general conditions.

Henri Lebesgue (1875–1941) in his 1902 doctoral thesis created a theory that solves both problems. His approach: measure the “size” of sets, rather than partitioning the domain into intervals.

Lebesgue Measure: A Generalization of Length

Let’s start simple: the length of the interval $[a,b]$ is $b - a$. The length of a finite union of non-overlapping intervals equals the sum of lengths (additivity). Can this “measurement” be extended to arbitrary sets?

Lebesgue Outer Measure: $m^*(E) = \inf{ \sum_k |I_k| : E \subseteq \bigcup_k I_k }$, where $I_k$ are intervals. We take the minimally possible total coverage of $E$ by a countable number of intervals.

A set $E$ is Lebesgue measurable (by the Carathéodory condition) if for any set $A$: $m^(A) = m^(A \cap E) + m^(A \cap E^{\mathsf{c}})$. Intuition: $E$ does not “splatter” irregularly but divides any test set cleanly. Then $m(E) = m^(E)$.

The $\Sigma$-algebra of measurable sets contains all Borel sets (open, closed, their countable unions/intersections). Countable additivity: if $E_k$ are pairwise disjoint, then $m(\bigcup E_k) = \sum m(E_k)$.

Key Examples:

• $m({x_0}) = 0$ for any point $x_0$.
• $m(\mathbb{Q} \cap [0,1]) = 0$: rational numbers form a countable set, which can be covered by intervals of total length $\varepsilon$ for any $\varepsilon > 0$.
• Cantor Set $C \subset [0,1]$: constructed by removing the middle thirds from $[0,1]$. $C$ is an uncountable closed perfect set without interior points. $m(C) = 0$! Uncountable but "measure zero."
• $m([0,1]) = 1$, $m((a,b)) = b - a$ regardless of whether the endpoints are open/closed.

The Lebesgue Integral: Horizontal Layers

Riemann builds the integral “vertically”: he partitions the $x$-axis and sums $f(x_k)\cdot \Delta x_k$. Lebesgue builds it “horizontally”: he partitions the $y$-value axis and, for each layer $[y_k, y_{k+1}]$, measures the "width" of the set ${x: f(x) \in [y_k, y_{k+1}]}$.

Step 1: Simple Functions. A function $\varphi = \sum_i c_i 1_{E_i}$ (takes a finite set of values $c_i$ on measurable sets $E_i$). The integral: $\int \varphi, d\mu = \sum_i c_i, m(E_i)$.

For $D(x)$: $D = 1 \cdot 1_{\mathbb{Q} \cap [0,1]} + 0 \cdot 1_{(\mathbb{R} \setminus \mathbb{Q}) \cap [0,1]}$. $\int D = 1 \cdot 0 + 0 \cdot 1 = 0$. ✓

Step 2: Non-Negative Measurable Functions. $\int f, d\mu = \sup { \int \varphi, d\mu: 0 \leq \varphi \leq f,\ \varphi\ \text{simple} }$.

Step 3: Functions of Changing Sign. Write $f = f^+ - f^-$, where $f^+ = \max(f,0) \geq 0$, $f^- = \max(-f,0) \geq 0$. If both integrals are finite: $\int f = \int f^+ - \int f^-$.

A function $f$ is Lebesgue integrable on $E$ ($f \in L^1(E)$) if and only if $\int_E |f| d\mu < \infty$.

Limit Theorems

This is the main practical advantage of Lebesgue over Riemann.

Levi’s Monotone Convergence Theorem (B. Levi, 1906): If $0 \leq f_1 \leq f_2 \leq \dots$ and $f_n \to f$ almost everywhere, then $\lim \int f_n = \int f$ (even if $\int f = +\infty$).

Fatou’s Lemma: If $f_n \geq 0$, then $\int(\liminf f_n) \leq \liminf \int f_n$.

Lebesgue’s Dominated Convergence Theorem (1904): If $f_n \to f$ almost everywhere and $|f_n(x)| \leq g(x)$ for some integrable $g$ ($\int g < \infty$), then:

$\int |f_n - f| \to 0$ and $\int f_n \to \int f$.

This is a powerful tool: it suffices to find a single summable “majorant” $g$, and the limit and the integral can be interchanged.

Example: $f_n(x) = n x e^{-n x}$ on $[0, \infty)$. $f_n \to 0$ for each $x > 0$. Find $\lim \int_0^\infty f_n(x) dx$.

Substitution: $\int_0^\infty n x e^{-n x} dx = (1/n)$ [by the formula $\int_0^\infty u e^{-u} du = 1$] $\cdot n \cdot (1/n) = 1.$ Limit $= 1 \neq 0 = \int \lim f_n$!

Here dominated convergence does not apply: there is no integrable majorant. The sequence “escapes to infinity”—the “mass” shifts and vanishes in the limit.

Riemann Integrability Criterion

Lebesgue’s Theorem (1901): A bounded function $f: [a,b] \to \mathbb{R}$ is Riemann integrable if and only if the set of its discontinuities has measure zero.

This explains why monotonic functions (the set of discontinuities is countable, measure $= 0$) are Riemann integrable, but the Dirichlet function is not.

If $f$ is Riemann integrable, it is Lebesgue integrable and the two integrals coincide. The Lebesgue integral is a strict extension of the Riemann integral.

Applications: Probability and Functional Analysis

The Lebesgue integral is foundational to modern probability theory (Kolmogorov, 1933). A probability space $(\Omega, F, P)$: $\Omega$—the space of elementary outcomes, $F$—a $\sigma$-algebra of events, $P$—a probability measure. The mathematical expectation of a random variable $X$: $E[X] = \int_\Omega X(\omega) dP(\omega)$—the Lebesgue integral.

The central limit theorem, law of large numbers, Radon–Nikodym theorem (conditional expectation)—all require the language of measure theory and the Lebesgue integral.

Question for Reflection: Show that the Thomae function $t(x) = 1/q$ if $x = p/q$ (irreducible fraction), and $t(x) = 0$ for irrational $x$, is Riemann integrable. What is its integral? How can you explain this using the Lebesgue theorem on the Riemann criterion?