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Lebesgue Integration

For a non-negative measurable simple function s=∑i=1naiχAis = \sum_{i=1}^n a_i \chi_{A_i} with ai≥0a_i \geq 0 and {Ai}\{A_i\} disjoint, define:

∫Xs dμ=∑i=1naiμ(Ai)\int_X s\, d\mu = \sum_{i=1}^n a_i \mu(A_i)

For a non-negative measurable function ff, define:

∫Xf dμ=sup⁡{∫Xs dμ:0≤s≤f, s simple}\int_X f\, d\mu = \sup\left\{\int_X s\, d\mu : 0 \leq s \leq f,\ s \text{ simple}\right\}

This definition is consistent with Theorem 5.4: by monotone convergence, we also have

∫Xf dμ=lim⁡n→∞∫Xsn dμ\int_X f\, d\mu = \lim_{n \to \infty} \int_X s_n\, d\mu

for any increasing sequence of simple functions sn↗fs_n \nearrow f.

For a measurable function f:X→Rf : X \to \mathbb{R}, define f+=max⁡(f,0)f^+ = \max(f, 0) and f−=max⁡(−f,0)f^- = \max(-f, 0), so f=f+−f−f = f^+ - f^- and ∣f∣=f++f−|f| = f^+ + f^-. If ∫f+ dμ<∞\int f^+\, d\mu < \infty and ∫f− dμ<∞\int f^-\, d\mu < \infty (i.e., ∫∣f∣ dμ<∞\int |f|\, d\mu < \infty), define:

∫Xf dμ=∫Xf+ dμ−∫Xf− dμ\int_X f\, d\mu = \int_X f^+\, d\mu - \int_X f^-\, d\mu

The function ff is called integrable (or f∈L1(μ)f \in L^1(\mu)) if ∫∣f∣ dμ<∞\int |f|\, d\mu < \infty.

Proposition 6.1 (Linearity). If f,g∈L1(μ)f, g \in L^1(\mu) and a,b∈Ra, b \in \mathbb{R}, then af+bg∈L1(μ)af + bg \in L^1(\mu) and ∫(af+bg) dμ=a∫f dμ+b∫g dμ\int(af + bg)\, d\mu = a\int f\, d\mu + b\int g\, d\mu.

Proposition 6.2 (Monotonicity). If f≤gf \leq g a.e., then ∫f dμ≤∫g dμ\int f\, d\mu \leq \int g\, d\mu.

Proposition 6.3 (Markov’s Inequality). If f≥0f \geq 0 is measurable, then for any a>0a > 0:

μ({x:∣f(x)∣≥a})≤1a∫∣f∣ dμ\mu(\{x : |f(x)| \geq a\}) \leq \frac{1}{a}\int |f|\, d\mu

Theorem 6.4 (Chebyshev’s Inequality). If f∈L2(μ)f \in L^2(\mu), then for any a>0a > 0:

μ({∣f−∫f dμ∣≥a})≤1a2Var(f)\mu(\{|f - \int f\, d\mu| \geq a\}) \leq \frac{1}{a^2}\mathrm{Var}(f)

Theorem 6.5 (Monotone Convergence Theorem — Levi). If 0≤f1≤f2≤⋯0 \leq f_1 \leq f_2 \leq \cdots are measurable and fn→ff_n \to f pointwise, then:

lim⁡n→∞∫fn dμ=∫f dμ\lim_{n \to \infty} \int f_n\, d\mu = \int f\, d\mu

Proof sketch. Let ss be a simple function with s≤fs \leq f. Define En={x:fn(x)≥(1−ε)s(x)}E_n = \{x : f_n(x) \geq (1 - \varepsilon)s(x)\}. Then En↗XE_n \nearrow X and ∫fn dμ≥(1−ε)∫s dμ\int f_n\, d\mu \geq (1 - \varepsilon)\int s\, d\mu for large nn. Take sup⁡\sup over ss and let ε→0\varepsilon \to 0. ■\blacksquare

Theorem 6.6 (Fatou’s Lemma). If fn≥0f_n \geq 0 are measurable, then:

∫lim inf⁡n→∞fn dμ≤lim inf⁡n→∞∫fn dμ\int \liminf_{n\to\infty} f_n\, d\mu \leq \liminf_{n\to\infty} \int f_n\, d\mu

Proof. Define gn=inf⁡k≥nfkg_n = \inf_{k \geq n} f_k. Then 0≤g1≤g2≤⋯0 \leq g_1 \leq g_2 \leq \cdots and gn→lim inf⁡fng_n \to \liminf f_n. By monotone convergence:

∫lim inf⁡fn dμ=lim⁡n→∞∫gn dμ≤lim inf⁡n→∞∫fn dμ\int \liminf f_n\, d\mu = \lim_{n\to\infty} \int g_n\, d\mu \leq \liminf_{n\to\infty} \int f_n\, d\mu

■\blacksquare

Theorem 6.7 (Dominated Convergence Theorem). If fn→ff_n \to f a.e. and there exists g∈L1(μ)g \in L^1(\mu) with ∣fn∣≤g|f_n| \leq g a.e. for all nn, then:

lim⁡n→∞∫fn dμ=∫f dμ\lim_{n\to\infty} \int f_n\, d\mu = \int f\, d\mu

Proof sketch. Apply Fatou’s lemma to g+fng + f_n and g−fng - f_n:

∫f dμ≤lim inf⁡∫fn dμand∫f dμ≥lim sup⁡∫fn dμ\int f\, d\mu \leq \liminf \int f_n\, d\mu \quad \text{and} \quad \int f\, d\mu \geq \limsup \int f_n\, d\mu

Combining gives the result. ■\blacksquare

Problem. Compute lim⁡n→∞∫01(1−x2/n)n dx\lim_{n \to \infty} \int_0^1 (1 - x^2/n)^n\, dx.

Solution. For each x∈[0,1]x \in [0, 1], (1−x2/n)n→e−x2(1 - x^2/n)^n \to e^{-x^2} as n→∞n \to \infty. Since 0≤(1−x2/n)n≤10 \leq (1 - x^2/n)^n \leq 1 for all nn and xx, we can apply the dominated convergence theorem with g(x)=1∈L1([0,1])g(x) = 1 \in L^1([0, 1]):

lim⁡n→∞∫01(1−x2/n)n dx=∫01e−x2 dx=π2 erf(1)≈0.7468\lim_{n\to\infty} \int_0^1 (1 - x^2/n)^n\, dx = \int_0^1 e^{-x^2}\, dx = \frac{\sqrt{\pi}}{2}\,\mathrm{erf}(1) \approx 0.7468

■\blacksquare

TheoremHypothesisConclusionRole
MCT (Levi)fn↗ff_n \nearrow f pointwise, fn≥0f_n \geq 0∫fn→∫f\int f_n \to \int fFoundation for all limit theorems
Fatou’s lemmafn≥0f_n \geq 0 measurable∫lim inf⁡fn≤lim inf⁡∫fn\int \liminf f_n \leq \liminf \int f_nWorks without convergence
DCTfn→ff_n \to f a.e., ∥fn∥≤g∈L1\|f_n\| \leq g \in L^1∫fn→∫f\int f_n \to \int fMost widely used limit theorem
Markov’s inequalityf≥0f \geq 0 meas., a>0a > 0μ({f≥a})≤1a∫f\mu(\{f \geq a\}) \leq \frac{1}{a}\int fBounds tail probabilities

The three convergence theorems are related: DCT follows from Fatou, and Fatou follows from MCT. Together they form the backbone of Lebesgue integration theory.

  • Applying DCT without a dominating function. If ∣fn∣≤g|f_n| \leq g fails for some nn, the limit may not pass through the integral. Fix: Always verify existence of g∈L1g \in L^1 dominating all fnf_n almost everywhere.
  • Confusing pointwise and uniform convergence. DCT only requires a.e. pointwise convergence, not uniform. Fix: The theorem is powerful precisely because it relaxes the uniform-convergence requirement of Riemann integration.
  • Assuming monotone convergence needs boundedness. MCT requires only monotonicity and non-negativity; the limit may be infinite. Fix: If ∫fn\int f_n diverges, the theorem correctly gives ∫f=∞\int f = \infty.
  • Forgetting the non-negativity in Fatou. Without fn≥0f_n \geq 0, the inequality ∫lim inf⁡fn≤lim inf⁡∫fn\int \liminf f_n \leq \liminf \int f_n can fail. Fix: Apply Fatou to fn+gf_n + g with gg integrable, then subtract.

6.8 Intuition: What Is Lebesgue Integration?

Section titled “6.8 Intuition: What Is Lebesgue Integration?”

Lebesgue integration is a more flexible way to define the integral that overcomes limitations of the Riemann integral. The Riemann integral partitions the domain (the xx-axis) into small intervals and sums the function values on each interval. The Lebesgue integral partitions the range (the yy-axis) and measures how much of the domain maps to each range interval. This swap of perspective is what makes Lebesgue integration more powerful.

The key advantage is that the Lebesgue integral can handle functions with wild discontinuities. The Riemann integral of the Dirichlet function (1 on rationals, 0 on irrationals) does not exist, but its Lebesgue integral is 0 because the rationals have measure zero. The monotone convergence theorem and dominated convergence theorem allow limits and integrals to be interchanged under very general conditions, which is essential for analysis. These theorems fail for Riemann integration, which requires uniform convergence. Lebesgue integration also provides the natural setting for LpL^p spaces, Fourier analysis, and probability theory.

  • Fourier series: DCT justifies term-by-term integration of Fourier series, allowing computation of coefficients by integrating the series.
  • Probability theory: Markov’s and Chebyshev’s inequalities are essential for proving laws of large numbers and concentration bounds.
  • LpL^p spaces: MCT and DCT are used to prove completeness of LpL^p spaces and to exchange limits with norms.
  • Fubini’s theorem: Tonelli’s theorem (MCT for non-negative functions) and Fubini’s theorem (DCT for integrable functions) justify swapping the order of integration.
Integral typeDefinitionKey property
Simple function∑aiχAi\sum a_i \chi_{A_i} with ai≥0a_i \geq 0, {Ai}\{A_i\} disjoint∫=∑aiμ(Ai)\int = \sum a_i \mu(A_i)
Non-negative meas.sup⁡{∫s:0≤s≤f, s simple}\sup\{\int s : 0 \leq s \leq f,\ s\text{ simple}\}MCT applies
General measurable∫f=∫f+−∫f−\int f = \int f^+ - \int f^-f∈L1f \in L^1 iff ∫∥f∥<∞\int\|f\| < \infty

6.10 Worked Example: Applying DCT to a Sequence with Oscillations

Section titled “6.10 Worked Example: Applying DCT to a Sequence with Oscillations”

Problem. Evaluate lim⁡n→∞∫0πsin⁡(nx)n dx\lim_{n\to\infty} \int_0^\pi \frac{\sin(nx)}{n}\, dx using the dominated convergence theorem.

Solution. Let fn(x)=sin⁡(nx)/nf_n(x) = \sin(nx)/n. For each x∈[0,π]x \in [0,\pi], ∣fn(x)∣≤1/n→0|f_n(x)| \leq 1/n \to 0, so fn→0f_n \to 0 pointwise. Also ∣fn(x)∣≤1|f_n(x)| \leq 1 for all nn and xx, and g(x)=1g(x) = 1 is integrable on [0,π][0,\pi]. By DCT:

lim⁡n→∞∫0πsin⁡(nx)n dx=∫0π0 dx=0\lim_{n\to\infty} \int_0^\pi \frac{\sin(nx)}{n}\, dx = \int_0^\pi 0\, dx = 0

We can verify directly: ∫0πsin⁡(nx) dx=[−cos⁡(nx)/n]0π=(1−(−1)n)/n\int_0^\pi \sin(nx)\, dx = [-\cos(nx)/n]_0^\pi = (1 - (-1)^n)/n, so the integral is 00 for even nn and 2/n22/n^2 for odd nn, both vanishing as n→∞n\to\infty.

■\blacksquare