The Lebesgue Integral and Convergence Theorems

⚡ Key Takeaways

One line: The Lebesgue integral extends Riemann integration by partitioning the range rather than the domain, yielding a theory in which limit and integral interchange under mild hypotheses (Monotone Convergence, Fatou, Dominated Convergence).

Key result in words: If a sequence of measurable functions converges pointwise and is suitably controlled (monotone, or dominated by an integrable function), then the integral of the limit equals the limit of the integrals.

Where it applies: Real and harmonic analysis, probability theory (expectation = integral against a probability measure), PDE, functional analysis ( $L^p$ spaces), ergodic theory.

Who/when: Henri Lebesgue (thesis 1902), building on Émile Borel's measure theory and Camille Jordan; convergence theorems refined by Pierre Fatou (1906), Beppo Levi (1906), Giuseppe Vitali, and later abstracted by Maurice Fréchet and Johann Radon.

💡 The Core Insight

Riemann slices the domain into intervals and asks whether the function varies little on each; this fails when $f$ oscillates wildly. Lebesgue slices the range into thin horizontal layers $[y_k, y_{k+1})$ and measures the size $\mu(f^{-1}[y_k,y_{k+1}))$ of the set where $f$ takes those values. Because measure is countably additive and stable under monotone limits, integration inherits exactly the limit-passage properties that Riemann's domain-partition lacks. Integration becomes measuring the region under the graph — but measured with the flexible, $\sigma$ -additive Lebesgue measure rather than the rigid Jordan content.

DefinitionMeasure space and measurable functions

A measure space is a triple $(X,\mathcal{M},\mu)$ where $\mathcal{M}$ is a $\sigma$ -algebra of subsets of $X$ and $\mu:\mathcal{M}\to[0,\infty]$ is countably additive:

1

\quad \mu\!\left(\bigsqcup_{n=1}^{\infty} E_n\right) = \sum_{n=1}^{\infty}\mu(E_n),\qquad E_n\in\mathcal{M}\text{ pairwise disjoint}.

Derivation hint: This is the defining axiom; finite additivity follows by padding with $\varnothing$ , and $\mu(\varnothing)=0$ is required to avoid $\mu\equiv\infty$ . A function $f:X\to[-\infty,\infty]$ is $\mathcal{M}$ -measurable if $f^{-1}((a,\infty])\in\mathcal{M}$ for every $a\in\mathbb{R}$ . Equivalently preimages of all Borel sets lie in $\mathcal{M}$ .

DefinitionThe Lebesgue integral, in three stages

Stage 1 — Simple functions. A simple function is $\varphi=\sum_{k=1}^{n} a_k \mathbf{1}_{A_k}$ with $a_k\ge 0$ , $A_k\in\mathcal{M}$ . Its integral is

2

\quad \int_X \varphi\,d\mu = \sum_{k=1}^{n} a_k\,\mu(A_k)\in[0,\infty],

Derivation hint: well-defined independent of representation because two simple representations share a common refinement on which finite additivity of $\mu$ forces agreement.

Stage 2 — Nonnegative measurable $f\ge 0$ :

3

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

Derivation hint: the supremum is taken over all simple minorants; monotonicity of the integral on simple functions makes this a genuine least-upper-bound construction approximating the area under the graph from below.

Stage 3 — General $f$ . Write $f=f^+-f^-$ with $f^+=\max(f,0)$ , $f^-=\max(-f,0)$ . If at least one of $\int f^+, \int f^-$ is finite, define

4

\quad \int_X f\,d\mu = \int_X f^+\,d\mu - \int_X f^-\,d\mu.

Derivation hint: the splitting is canonical since $f^+f^-=0$ ; we forbid the case $\infty-\infty$ , which is when $f$ is non-integrable. $f$ is <strong>integrable</strong> iff $\int|f|\,d\mu<\infty$ , i.e. both parts finite.

TheoremThe Three Convergence Theorems

Let $(X,\mathcal{M},\mu)$ be a measure space and $f_n:X\to[-\infty,\infty]$ measurable.

(MCT) Monotone Convergence (Beppo Levi). If $0\le f_1\le f_2\le\cdots$ and $f_n\uparrow f$ pointwise, then

5

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

Derivation hint: monotonicity gives " $\le$ "; the reverse uses that for any simple $\varphi\le f$ and $0<c<1$ , the sets $E_n=\{f_n\ge c\varphi\}$ increase to $X$ , then apply continuity of $\mu$ from below.

(Fatou) Fatou's Lemma. For $f_n\ge 0$ ,

6

\quad \int_X \liminf_{n\to\infty} f_n\,d\mu \le \liminf_{n\to\infty}\int_X f_n\,d\mu.

Derivation hint: apply MCT to $g_n=\inf_{k\ge n} f_k\uparrow \liminf f_n$ , and use $g_n\le f_n$ so $\int g_n\le\int f_n$ .

(DCT) Dominated Convergence (Lebesgue). If $f_n\to f$ pointwise (a.e.) and there exists integrable $g$ with $|f_n|\le g$ for all $n$ , then $f$ is integrable and

7

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

Derivation hint: apply Fatou to the nonnegative sequences $g+f_n$ and $g-f_n$ ; the two resulting inequalities sandwich $\int f$ between $\liminf$ and $\limsup$ of $\int f_n$ .

ProofMonotone Convergence Theorem

Step 1 (the easy inequality). Since $f_n\le f_{n+1}\le f$ , monotonicity of the integral on nonnegative functions gives $\int f_n\le \int f_{n+1}\le \int f$ . Hence $L:=\lim_n\int f_n$ exists in $[0,\infty]$ and $L\le \int f$ .

Step 2 (setup for the reverse). Fix a simple function $\varphi=\sum_{k} a_k\mathbf{1}_{A_k}$ with $0\le\varphi\le f$ , and fix a constant $c\in(0,1)$ . Define

E_n=\{x\in X: f_n(x)\ge c\,\varphi(x)\}\in\mathcal{M}.

Step 3 ( $E_n$ exhausts $X$ ). Because $f_n\uparrow f\ge\varphi> c\varphi$ wherever $\varphi>0$ (strict since $c<1$ ), and trivially where $\varphi=0$ , every $x$ eventually lies in $E_n$ . Thus $E_n\uparrow X$ .

Step 4 (estimate). On $E_n$ , $f_n\ge c\varphi$ , so

\int_X f_n\,d\mu \ge \int_{E_n} f_n\,d\mu \ge c\int_{E_n}\varphi\,d\mu = c\sum_k a_k\,\mu(A_k\cap E_n).

Step 5 (continuity from below). Since $A_k\cap E_n\uparrow A_k$ , $\sigma$ -additivity gives $\mu(A_k\cap E_n)\uparrow\mu(A_k)$ (Equation [1]). Letting $n\to\infty$ :

L\ge c\sum_k a_k\,\mu(A_k)=c\int_X\varphi\,d\mu.

Step 6 (remove $c$ and $\varphi$ ). Let $c\uparrow 1$ : $L\ge\int\varphi$ . Take the supremum over all simple $\varphi\le f$ : by definition (Equation [3]), $L\ge\int f$ . Combined with Step 1, $L=\int f$ . $\blacksquare$

∎

ProofDominated Convergence from Fatou

Step 1. Since $|f_n|\le g$ and $f_n\to f$ a.e., $|f|\le g$ a.e., so $f$ is integrable. Consider $h_n^\pm = g\pm f_n\ge 0$ .

Step 2. Apply Fatou (Equation [6]) to $h_n^+ = g+f_n$ :

\int(g+f) \le \liminf_n\int(g+f_n)=\int g+\liminf_n\int f_n.

Cancel the finite $\int g$ : $\;\int f\le \liminf_n\int f_n$ .

Step 3. Apply Fatou to $h_n^- = g-f_n$ :

\int(g-f)\le\liminf_n\int(g-f_n)=\int g-\limsup_n\int f_n.

Cancel $\int g$ : $\;\limsup_n\int f_n\le \int f$ .

Step 4. Combining: $\;\limsup_n\int f_n\le\int f\le\liminf_n\int f_n$ , forcing $\lim_n\int f_n=\int f$ .

Step 5 ( $L^1$ convergence). Apply the same argument to $|f_n-f|\le 2g$ with pointwise limit $0$ : $\int|f_n-f|\to 0$ . $\blacksquare$

∎

HL

Henri Léon Lebesgue

1875–1941

A French mathematician whose 1902 doctoral thesis Intégrale, longueur, aire at the Sorbonne overturned the foundations of integration. Lebesgue's signature insight — partitioning the range, "summing the money in your pocket by denomination rather than by the order you pull coins out" (a metaphor he himself used) — gave analysis its modern measure-theoretic backbone. He was initially met with skepticism; his contemporaries doubted the need to integrate such "pathological" functions. Lebesgue spent much of his career in pedagogy and held chairs at Rennes, Poitiers, and finally the Collège de France (1921). He was, somewhat ironically, suspicious of the abstract axiomatic generalizations (general measure spaces) his own work spawned, preferring the concrete geometry of the real line.

🧠 Intuition

Picture the graph of $f\ge 0$ over $X$ . Riemann drops vertical fences at points $x_0<x_1<\cdots$ and sums rectangle areas $f(\xi_i)\,(x_{i+1}-x_i)$ ; if $f$ jumps madly within a single $[x_i,x_{i+1}]$ , the choice of sample $\xi_i$ matters and the sum need not converge. Lebesgue instead draws horizontal lines at heights $y_0<y_1<\cdots$ , asks "how much of $X$ has value in $[y_k,y_{k+1})$ ?", and sums $y_k\cdot\mu(\{y_k\le f<y_{k+1}\})$ . The set $\{y_k\le f<y_{k+1}\}$ can be a hopelessly complicated set — a fractal, the irrationals, anything measurable — and Lebesgue measure handles it. This is why the convergence theorems work: monotone increasing $f_n$ correspond to the horizontal layers filling in from the bottom, and countable additivity (continuity from below) lets you take the limit layer by layer without losing any area.

Figure

Range-partition (Lebesgue, gold) vs domain-partition (Riemann, gray): Lebesgue measures the preimage set of each horizontal band.

ExampleThe Dirichlet function — Riemann fails, Lebesgue trivial

Given: $f=\mathbf{1}_{\mathbb{Q}\cap[0,1]}$ , the indicator of the rationals in $[0,1]$ , with Lebesgue measure $\lambda$ .

Step 1. $f$ is a simple function: $f=1\cdot\mathbf{1}_{A}$ with $A=\mathbb{Q}\cap[0,1]$ .

Step 2. Compute $\lambda(A)$ . The rationals are countable: $A=\{q_1,q_2,\dots\}$ . By countable additivity (Equation [1]) and $\lambda(\{q_k\})=0$ ,

\lambda(A)=\sum_{k=1}^\infty \lambda(\{q_k\})=\sum_{k=1}^\infty 0 = 0.

Step 3. By Equation [2], $\int_{[0,1]} f\,d\lambda = 1\cdot\lambda(A)=1\cdot 0$ .

Boxed answer:

\boxed{\displaystyle\int_{[0,1]}\mathbf{1}_{\mathbb{Q}}\,d\lambda = 0}

Note: the Riemann integral does not exist — upper sums are $1$ , lower sums are $0$ , for every partition.

ExampleInterchanging limit and integral via DCT

Given: Evaluate $\displaystyle \lim_{n\to\infty}\int_0^1 \frac{n\,x}{1+n^2 x^2}\,dx$ . (Lebesgue measure on $[0,1]$ .)

Step 1 — pointwise limit. For fixed $x\in(0,1]$ , $f_n(x)=\dfrac{nx}{1+n^2x^2}=\dfrac{x/n}{1/n^2+x^2}\to 0$ as $n\to\infty$ . At $x=0$ , $f_n(0)=0$ . So $f_n\to 0$ pointwise.

Step 2 — find a dominating function. By AM–GM, $1+n^2x^2\ge 2nx$ , hence

|f_n(x)| = \frac{nx}{1+n^2x^2}\le \frac{nx}{2nx}=\frac{1}{2}=:g(x).

The constant $g=\tfrac12$ is integrable on $[0,1]$ since $\int_0^1 \tfrac12\,dx=\tfrac12<\infty$ .

Step 3 — apply DCT (Equation [7]). Domination $|f_n|\le g$ and $f_n\to 0$ give

\lim_{n\to\infty}\int_0^1 f_n\,dx = \int_0^1 \lim_{n\to\infty} f_n\,dx = \int_0^1 0\,dx.

Boxed answer:

\boxed{\displaystyle \lim_{n\to\infty}\int_0^1 \frac{n x}{1+n^2x^2}\,dx = 0}

(Sanity check: the exact integral is $\tfrac{1}{2n}\ln(1+n^2)\to 0$ , confirming DCT.)

⚠️ Domination is not optional

Without an integrable dominating function $g$ , the conclusion of DCT can fail even when $f_n\to 0$ pointwise everywhere and each $\int f_n$ exists. Pointwise convergence alone never guarantees interchange. Likewise, MCT requires the sequence to be monotone increasing and nonnegative — a decreasing sequence to $0$ can have $\int f_n\not\to 0$ (mass escapes to infinity). Fatou's inequality can be strict; equality is the exception, not the rule.

CounterexampleThe escaping bump — DCT and MCT both can fail without hypotheses

On $(\mathbb{R},\lambda)$ let $f_n=\mathbf{1}_{[n,n+1]}$ . Then $f_n\to 0$ pointwise (for each fixed $x$ , $f_n(x)=0$ once $n>x$ ), but

\int_{\mathbb{R}} f_n\,d\lambda = \lambda([n,n+1])=1 \ \not\to\ 0 = \int_{\mathbb{R}} 0\,d\lambda.

Fatou is satisfied (and strict): $\int\liminf f_n=0 < 1=\liminf\int f_n$ . DCT fails because any $g\ge \sup_n f_n=\mathbf{1}_{[1,\infty)}$ has $\int g=\infty$ — no integrable dominator exists. A second classic: the "tall thin spike" $g_n=n\,\mathbf{1}_{[0,1/n]}$ has $\int g_n=1$ for all $n$ but $g_n\to 0$ a.e.; mass escapes vertically. Both show: pointwise convergence controls nothing about integrals.

connection

The Lebesgue integral is the analytic engine behind several neighbouring theories. In probability, expectation $\mathbb{E}[X]=\int_\Omega X\,d\mathbb{P}$ is literally a Lebesgue integral against a probability measure; the convergence theorems become the Monotone/Dominated Convergence for expectations, and Fatou underlies many a.s.-convergence arguments. The completeness of Lp Spaces (Riesz–Fischer theorem) is proved precisely via the convergence theorems applied to rapidly Cauchy subsequences. Fubini–Tonelli extends single integrals to product measures, with Tonelli's half resting on MCT and Fubini's on DCT. The Radon-Nikodym Theorem characterizes when one measure is an integral of a density against another, generalizing the integral itself. And Differentiation Under the Integral Sign (Leibniz rule) is just DCT applied to difference quotients.

🔬 Where this appears

⏳ History

- 1854 — Bernhard Riemann, in his Habilitation, gives the first rigorous definition of the integral via domain partitions and the Riemann criterion.

- 1881–1884 — Vito Volterra and others construct functions (e.g. Volterra's function) exposing the limits of Riemann integration vis-à-vis the Fundamental Theorem of Calculus.

- 1892–1898 — Camille Jordan and Giuseppe Peano develop "Jordan content," a finitely-additive notion of measure foreshadowing but inadequate for limits.

- 1898 — Émile Borel introduces countably-additive measure on the Borel sets of

\mathbb{R}

in his Leçons sur la théorie des fonctions, the decisive conceptual step.

- 1902 — Henri Lebesgue's thesis Intégrale, longueur, aire defines Lebesgue measure and integral, proving the bounded convergence theorem.

- 1906 — Pierre Fatou's thesis on trigonometric series states and uses Fatou's Lemma; Beppo Levi proves the Monotone Convergence Theorem the same year.

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

L^2

(Riesz–Fischer theorem), revealing

L^p

as Banach spaces.

- 1910–1913 — Johann Radon and Maurice Fréchet abstract integration to general measure spaces; Radon measures and the Radon–Nikodym derivative emerge.

- 1933 — Andrey Kolmogorov's Grundbegriffe founds probability on measure theory, making the Lebesgue integral the definition of expectation.

open-problem

The Lebesgue integral is absolutely convergent — it integrates $f$ only when $\int|f|<\infty$ — and so cannot directly assign $\int_0^\infty \frac{\sin x}{x}\,dx=\frac{\pi}{2}$ (an improper, conditionally convergent Riemann integral). This motivates gauge / Henstock–Kurzweil integration, which recovers all such conditionally convergent integrals and contains Lebesgue's theory as the absolutely-integrable subcase; the precise functional-analytic structure of the Henstock space remains less developed than $L^p$ theory. A second frontier is non-measurable phenomena: assuming the Axiom of Choice, Vitali sets and Banach–Tarski paradoxes show Lebesgue measure cannot be extended to all subsets of $\mathbb{R}^n$ translation-invariantly; the consistency strength of "every set is Lebesgue measurable" (Solovay's model, requiring an inaccessible cardinal) ties measure theory to large-cardinal axioms. Finally, infinite-dimensional analogues fail: there is no translation-invariant $\sigma$ -finite measure on an infinite-dimensional Banach space, forcing the use of Gaussian (Wiener) measures and abstract Wiener spaces, where the right convergence theorems are still actively studied.

The Lebesgue Integral and Convergence Theorems

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also