The Lebesgue Integral Introduction

⚡ Key Takeaways

One-line statement: The Lebesgue integral measures the "area under a curve" by partitioning the range (y-axis values) rather than the domain (x-axis), enabling integration of far more functions than the Riemann integral. Key result in words: A function is Lebesgue integrable if it can be approximated from below by simple (step-like) functions whose measured "horizontal slices" sum to a finite value, and the integral is the supremum of these sums. Where it applies: Probability theory, Fourier analysis, functional analysis, partial differential equations, and any setting requiring robust convergence theorems for sequences of functions. Who/when: Henri Lebesgue, doctoral thesis Intégrale, longueur, aire (1902), Paris.

💡 The Core Insight

Riemann partitions the domain; Lebesgue partitions the range. To integrate a non-negative function $f$ , fix a height-level $y$ and measure the set $\{x : f(x) > y\}$ . Stacking these measured sets across all heights — like accounting for a sum of money by sorting coins by denomination rather than by the order received — yields a quantity stable under limits. The price of admission is that we must first know how to measure arbitrary sets, which is the content of Lebesgue Measure.

DefinitionMeasurable Function

Let $(X, \mathcal{M}, \mu)$ be a measure space: $X$ a set, $\mathcal{M}$ a $\sigma$ -algebra of "measurable" subsets, and $\mu : \mathcal{M} \to [0,\infty]$ a measure. A function $f : X \to [-\infty, \infty]$ is measurable if for every $\alpha \in \mathbb{R}$ ,

1

\quad \{x \in X : f(x) > \alpha\} \in \mathcal{M}.

Derivation hint: The condition for all $\alpha$ is equivalent to preimages of all Borel sets being measurable; use that the half-lines $(\alpha, \infty]$ generate the Borel $\sigma$ -algebra on $[-\infty,\infty]$ . Here $\mathcal{M}$ is closed under complements and countable unions, and $\mu(\varnothing)=0$ with countable additivity. On $\mathbb{R}$ , $\mu$ is Lebesgue Measure and $\mathcal{M}$ the Lebesgue-measurable sets.

DefinitionSimple Function and Its Integral

A simple function is a measurable function taking finitely many values $a_1,\dots,a_n \ge 0$ , written

2

\quad s(x) = \sum_{i=1}^{n} a_i \,\mathbf{1}_{A_i}(x), \qquad A_i = \{x : s(x) = a_i\},

where $\mathbf{1}_{A_i}$ is the indicator function of the measurable set $A_i$ . Its Lebesgue integral over $E \in \mathcal{M}$ is

3

\quad \int_E s\,d\mu = \sum_{i=1}^{n} a_i\,\mu(A_i \cap E).

Derivation hint: This is the only definition consistent with linearity and with $\int_E \mathbf{1}_A\,d\mu = \mu(A\cap E)$ ; verify it does not depend on the chosen representation of $s$ by refining two partitions to a common one.

DefinitionThe Lebesgue Integral of a Non-negative Function

For a measurable $f : X \to [0,\infty]$ , define

4

\quad \int_E f\,d\mu = \sup\left\{ \int_E s\,d\mu : 0 \le s \le f,\ s \text{ simple}\right\}.

Derivation hint: The supremum is over all simple functions dominated by $f$ ; monotonicity of $\mu$ guarantees the set of values is bounded below by $0$ and the sup is well-defined in $[0,\infty]$ . For general measurable $f$ , split $f = f^+ - f^-$ where $f^+ = \max(f,0)$ and $f^- = \max(-f,0)$ . Then

5

\quad \int_E f\,d\mu = \int_E f^+\,d\mu - \int_E f^-\,d\mu,

provided not both terms are $+\infty$ . The function $f$ is Lebesgue integrable on $E$ iff

6

\quad \int_E |f|\,d\mu = \int_E f^+\,d\mu + \int_E f^-\,d\mu < \infty.

Derivation hint: Since $|f| = f^+ + f^-$ and both $f^\pm \le |f|$ , finiteness of $\int|f|$ forces finiteness of each piece; the space of such functions is denoted $L^1(\mu)$ .

TheoremMonotone Convergence Theorem (MCT)

If $\{f_n\}$ is a sequence of measurable functions with $0 \le f_1 \le f_2 \le \cdots$ and $f_n \to f$ pointwise, then

7

\quad \lim_{n\to\infty} \int_E f_n\,d\mu = \int_E f\,d\mu = \int_E \lim_{n\to\infty} f_n\,d\mu.

Derivation hint: The limit and supremum can be exchanged; prove $\ge$ trivially from monotonicity, and $\le$ by approximating $f$ from below with a simple $s$ scaled by $c<1$ and the sets $E_n = \{f_n \ge cs\}$ increasing to $E$ .

TheoremDominated Convergence Theorem (DCT)

Let $f_n \to f$ pointwise a.e., and suppose there exists an integrable $g$ with $|f_n| \le g$ for all $n$ . Then $f$ is integrable and

8

\quad \lim_{n\to\infty}\int_E f_n\,d\mu = \int_E f\,d\mu, \qquad \lim_{n\to\infty}\int_E |f_n - f|\,d\mu = 0.

Derivation hint: Apply Fatou's Lemma to the non-negative sequences $g + f_n$ and $g - f_n$ ; the dominating $g$ controls both directions of the inequality, sandwiching the limit.

ProofConstruction of the Integral via the Simple Function Approximation Theorem

We justify that the supremum in [4] is genuinely computable by exhibiting an increasing sequence of simple functions converging to any non-negative measurable $f$ .

Step 1 — Dyadic slicing of the range. For each $n \in \mathbb{N}$ , partition $[0, 2^n)$ into bands of height $2^{-n}$ and cap at $2^n$ . Define

9

\quad s_n(x) = \begin{cases} \dfrac{k-1}{2^n} & \text{if } \dfrac{k-1}{2^n} \le f(x) < \dfrac{k}{2^n},\ k=1,\dots,n2^n,\\[2mm] n & \text{if } f(x) \ge n. \end{cases}

Derivation hint: Each set $\{(k-1)/2^n \le f < k/2^n\}$ is measurable as the difference of two superlevel sets from [1], so $s_n$ is a legitimate simple function.

Step 2 — Monotonicity. Refining $n \to n+1$ halves each band, so $s_n \le s_{n+1}$ everywhere. The approximation always undershoots: $s_n \le f$ .

Step 3 — Pointwise convergence. If $f(x) < \infty$ , then for $n > f(x)$ we have $0 \le f(x) - s_n(x) < 2^{-n} \to 0$ . If $f(x) = \infty$ , then $s_n(x) = n \to \infty = f(x)$ .

Step 4 — Integral via MCT. By the Monotone Convergence Theorem [7],

10

\quad \int_E f\,d\mu = \lim_{n\to\infty} \int_E s_n\,d\mu = \lim_{n\to\infty}\sum_{k} \frac{k-1}{2^n}\,\mu\!\left(\left\{\tfrac{k-1}{2^n}\le f < \tfrac{k}{2^n}\right\}\cap E\right).

Derivation hint: Each $\int_E s_n\,d\mu$ is a finite sum computed by [3]; the limit exists in $[0,\infty]$ because the sequence of integrals is increasing. This proves the supremum in [4] is attained as a limit, and gives a concrete recipe: the Lebesgue integral is the limit of "height-band measure" sums. $\blacksquare$

∎

HL

Henri Léon Lebesgue

1875–1941

Lebesgue, a student at the École Normale Supérieure, revolutionized analysis with his 1902 thesis under the indirect influence of Émile Borel's measure theory and Camille Jordan's content. His insight — "I have to pay a sum; I empty my pockets and put coins and bills on the table by value, then sum" — captures the range-partition philosophy. He resolved long-standing pathologies (term-by-term integration of series, the fundamental theorem of calculus for absolutely continuous functions) and gave precise meaning to "almost everywhere." His name now attaches to measure, integral, dominated convergence, the Lebesgue differentiation theorem, and $L^p$ spaces.

🧠 Intuition

Picture earning income over a year and wanting your total. The Riemann accountant goes day by day (slicing the time-domain), adding each day's receipts. The Lebesgue accountant instead asks: "On how many days did I earn between $0 and $10? Between $10 and $20?" — then multiplies each income-band by the number of days it occurred. Both reach the same total when both methods work. But if your daily income is wildly erratic (jumping between a fixed set of values in no orderly pattern), the day-by-day Riemann sums refuse to settle, while the Lebesgue band-by-band sorting still produces a clean answer. The Lebesgue method is robust precisely because measure doesn't care about the arrangement of points, only about how much total "stuff" lands in each band.

Figure

Riemann vertical strips (domain partition) versus Lebesgue horizontal bands (range partition)

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ExampleThe Dirichlet Function — Lebesgue Integrable, Not Riemann

Given: $f : [0,1] \to \mathbb{R}$ defined by $f(x)=1$ if $x \in \mathbb{Q}$ , $f(x)=0$ if $x \notin \mathbb{Q}$ . Use Lebesgue measure $\mu = m$ .

Step 1 — Riemann fails. Every subinterval contains both rationals and irrationals, so the upper Darboux sum is $1$ and the lower is $0$ for every partition. They never agree, so $f$ is not Riemann integrable.

Step 2 — Identify the level sets. $f$ is already simple: $f = 1\cdot \mathbf{1}_{\mathbb{Q}\cap[0,1]} + 0\cdot\mathbf{1}_{(\mathbb{Q}^c)\cap[0,1]}$ .

Step 3 — Measure the sets. The rationals are countable, hence $m(\mathbb{Q}\cap[0,1]) = 0$ . Therefore $m(\text{irrationals}\cap[0,1]) = 1$ .

Step 4 — Apply [3].

\int_{[0,1]} f\,dm = 1\cdot m(\mathbb{Q}\cap[0,1]) + 0\cdot m(\text{irrationals}) = 1\cdot 0 + 0\cdot 1.

\boxed{\displaystyle\int_{[0,1]} f\,dm = 0 \quad (\text{dimensionless})}

ExampleInterchanging Limit and Integral via MCT

Given: $f_n(x) = \dfrac{n x}{1 + n^2 x^2}\cdot \mathbf{1}_{[0,1]}(x)$ ? — instead take the cleaner monotone case $f_n(x) = x^{1/n}$ on $[0,1]$ , increasing to $f(x)=1$ for $x\in(0,1]$ . Compute $\lim_n \int_0^1 x^{1/n}\,dm$ .

Step 1 — Verify hypotheses. For fixed $x\in(0,1)$ , $x^{1/n}$ increases toward $x^0 = 1$ as $n\to\infty$ (since $1/n \downarrow 0$ and $0<x<1$ ). Each $f_n \ge 0$ and measurable (continuous). MCT [7] applies.

Step 2 — Compute each integral (Lebesgue agrees with Riemann for continuous functions):

\int_0^1 x^{1/n}\,dm = \left[\frac{x^{1/n + 1}}{1/n + 1}\right]_0^1 = \frac{1}{1 + 1/n} = \frac{n}{n+1}.

Step 3 — Take the limit.

\lim_{n\to\infty}\int_0^1 f_n\,dm = \lim_{n\to\infty}\frac{n}{n+1} = 1.

Step 4 — Confirm via MCT. $\int_0^1 f\,dm = \int_0^1 1\,dm = m((0,1]) = 1$ . The two match exactly, as guaranteed.

\boxed{\displaystyle\lim_{n\to\infty}\int_0^1 x^{1/n}\,dm = 1 \quad (\text{dimensionless})}

⚠️ Common mistake

"Every Riemann integrable function is Lebesgue integrable, and vice versa." Only the first half is true on a bounded interval: if $f$ is Riemann integrable on $[a,b]$ , it is Lebesgue integrable with the same value. The converse fails — the Dirichlet function above is Lebesgue but not Riemann integrable. For improper integrals the relationship inverts in subtle ways: $\int_0^\infty \frac{\sin x}{x}\,dx$ converges as an improper Riemann integral (to $\pi/2$ ) but is not Lebesgue integrable, because $\int_0^\infty \left|\frac{\sin x}{x}\right|\,dx = \infty$ and the Lebesgue integral requires absolute integrability per [6].

CounterexampleWhy Boundedness of

|f_n|

Alone Is Insufficient — the Moving Bump

Let $f_n(x) = n\cdot\mathbf{1}_{(0, 1/n)}(x)$ on $[0,1]$ . Then $f_n(x)\to 0$ pointwise (for any fixed $x>0$ , eventually $1/n < x$ , and $f_n(0)=0$ always). So $\int_0^1 \lim_n f_n\,dm = \int_0^1 0\,dm = 0$ . But

11

\quad \int_0^1 f_n\,dm = n\cdot m\!\left(\left(0,\tfrac1n\right)\right) = n\cdot\tfrac1n = 1 \quad \text{for every } n.

Derivation hint: Use [3] with the single nonzero value $a_1=n$ on the set $A_1=(0,1/n)$ of measure $1/n$ . Thus $\lim_n \int f_n = 1 \neq 0 = \int \lim_n f_n$ . The interchange fails because there is no integrable dominating function $g$ : any $g \ge \sup_n f_n$ would need $g(x) \ge n$ near $0$ , forcing $\int g = \infty$ . This is exactly why DCT [8] requires a dominating envelope — MCT does not apply either since $\{f_n\}$ is not monotone.

connection

The Lebesgue integral sits at the crossroads of several theories. It is built atop Lebesgue Measure and the abstract Measure Theory of $\sigma$ -algebras. Its convergence theorems power Lp Spaces, the complete normed spaces $L^p(\mu)$ central to Functional Analysis. In probability, $\int_X X\,dP$ is the expected value — Probability Theory is literally measure theory with $\mu(X)=1$ . The integral also clarifies the Fundamental Theorem of Calculus via absolutely continuous functions and the Lebesgue differentiation theorem, and it provides the natural setting for Fourier Analysis where Parseval's identity demands $L^2$ completeness that the Riemann integral cannot supply.

🔬 Where this appears

⏳ History

- 1854 — Bernhard Riemann formalizes the Riemann integral in his Habilitationsschrift, the first rigorous definition of the integral.

- 1881–1884 — Vito Volterra constructs a bounded derivative that is not Riemann integrable, exposing limits of the Riemann theory.

- 1898 — Émile Borel develops the theory of measure for "Borel sets," assigning length consistently to a large class of subsets.

- 1901 — Lebesgue announces his integral in a Comptes Rendus note, "Sur une généralisation de l'intégrale définie."

- 1902 — Lebesgue's doctoral thesis Intégrale, longueur, aire establishes Lebesgue measure and the integral, including the dominated and monotone convergence ideas.

- 1906 — Pierre Fatou proves Fatou's Lemma in his study of analytic functions and trigonometric series.

- 1907 — Frigyes Riesz and Ernst Fischer prove the completeness of

L^2

(Riesz–Fischer theorem), cementing the integral's role in Fourier theory.

- 1910 — Lebesgue's Sur l'intégration des fonctions discontinues extends the theory to higher dimensions and differentiation of the integral.

- 1918–1933 — Maurice Fréchet, Constantin Carathéodory (1918, outer-measure construction), and Andrey Kolmogorov (1933, Grundbegriffe) abstract the theory into general measure spaces and axiomatic probability.

open-problem

Non-measurable sets and the limits of "measuring everything." Using the Axiom of Choice, Giuseppe Vitali (1905) constructed a subset of $[0,1]$ that is not Lebesgue measurable — so [1] genuinely restricts which functions we can integrate. The Banach–Tarski paradox (1924) sharpens this: in $\mathbb{R}^3$ a ball can be decomposed into finitely many (necessarily non-measurable) pieces and reassembled into two balls, showing finitely-additive volume cannot extend to all sets in a rotation-invariant way. Robert Solovay (1970) proved that, consistent with ZF (without full Choice), one can have a model where every set of reals is Lebesgue measurable — so the existence of non-measurable sets is intimately tied to the strength of the axioms one accepts. Open meta-questions remain about which extensions of measure (e.g., to non- $\sigma$ -finite or non-locally-compact settings) preserve the convergence theorems, and how Lebesgue integration interacts with constructive/computable analysis where the supremum in [4] may not be effectively attainable.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also