Lebesgue Convergence Theorems

Monotone convergence, Fatou, dominated convergence — the three pillars for swapping limits and integrals. Watch mass escape (escape·spike) to see why each hypothesis is needed.

fn↑f⇒∫fn↑∫ff_n \uparrow f \Rightarrow \int f_n \uparrow \int f
∫lim inf⁡fn≤lim inf⁡∫fn\int \liminf f_n \le \liminf \int f_n
∣fn∣≤g, fn→f a.e. ⇒∫fn→∫f|f_n| \le g,\ f_n \to f \text{ a.e. } \Rightarrow \int f_n \to \int f

Examples & Exercises

For the spike sequence fn=n 1[0,1/n]f_n = n\,\mathbf{1}_{[0,1/n]}, compute both sides of Fatou's lemma, ∫lim inf⁡fn\int \liminf f_n and lim inf⁡∫fn\liminf \int f_n, directly and show the inequality is strict.

∫fn=n⋅1n=1∀n\int f_n = n \cdot \frac{1}{n} = 1 \quad \forall n
∫lim inf⁡n→∞fn=0<1=lim inf⁡n→∞∫fn\int \liminf_{n \to \infty} f_n = 0 < 1 = \liminf_{n \to \infty} \int f_n
f10=10 1[0,1/10]f_{10} = 10\,\mathbf{1}_{[0,1/10]}: ∫f10=1\int f_{10} = 1 yet the pointwise limit is 00, so ∫lim inf⁡fn=0\int \liminf f_n = 0 — Fatou's inequality is strict.
Show proof
  1. Each fnf_n is a step function equal to nn on [0,1/n][0, 1/n] and 00 elsewhere, so ∫fn=n⋅m([0,1/n])=n⋅1n=1\int f_n = n \cdot m([0, 1/n]) = n \cdot \frac{1}{n} = 1 — the same for every nn. Hence lim inf⁡n→∞∫fn=1\liminf_{n \to \infty} \int f_n = 1.
  2. Now the pointwise lim inf⁡\liminf. For x>0x > 0, once n>1/xn > 1/x we have 1/n<x1/n < x and fn(x)=0f_n(x) = 0 — so lim inf⁡nfn(x)=0\liminf_n f_n(x) = 0. At x=0x = 0, fn(0)=n→∞f_n(0) = n \to \infty, so the lim inf⁡\liminf is ∞\infty there, but a single point is a null set and does not affect the integral. Thus lim inf⁡nfn=0\liminf_n f_n = 0 a.e. and ∫lim inf⁡nfn=0\int \liminf_n f_n = 0.
  3. Hence 0=∫lim inf⁡fn<lim inf⁡∫fn=10 = \int \liminf f_n < \liminf \int f_n = 1 — Fatou's inequality ∫lim inf⁡fn≤lim inf⁡∫fn\int \liminf f_n \le \liminf \int f_n holds as a strict inequality. The mass concentrated at a single point and vanished in the limit. This proof is for this concrete sequence: it verifies the general Fatou lemma on a fully computable case rather than proving the general theorem.

Show the sequence fn=1[0,1−1/n]f_n = \mathbf{1}_{[0,1-1/n]} increases to f=1[0,1)f = \mathbf{1}_{[0,1)}, and prove ∫fn↑∫f\int f_n \uparrow \int f with explicit numbers.

0≤fn≤fn+1,fn(x)↑1[0,1)(x)0 \le f_n \le f_{n+1}, \quad f_n(x) \uparrow \mathbf{1}_{[0,1)}(x)
∫fn=1−1n↑1=∫lim⁡fn\int f_n = 1 - \frac{1}{n} \uparrow 1 = \int \lim f_n
n=10n = 10: ∫f10=1−110=0.9\int f_{10} = 1 - \frac{1}{10} = 0.9 increases toward 11, and the pointwise limit 1[0,1)\mathbf{1}_{[0,1)} integrates to 11.
Show proof
  1. Since 1−1n1 - \frac{1}{n} increases, [0,1−1n]⊂[0,1−1n+1][0, 1-\frac{1}{n}] \subset [0, 1-\frac{1}{n+1}], hence 0≤fn≤fn+10 \le f_n \le f_{n+1} — a monotone increasing sequence of nonnegative measurable functions.
  2. Pointwise limit: if x<1x < 1, eventually 1−1n>x1 - \frac{1}{n} > x, so fn(x)=1f_n(x) = 1 and the limit is 11. If x≥1x \ge 1, fn(x)=0f_n(x) = 0 for all nn, so the limit is 00. That is, fn↑1[0,1)f_n \uparrow \mathbf{1}_{[0,1)} pointwise.
  3. ∫fn=m([0,1−1n])=1−1n↑1\int f_n = m([0, 1-\frac{1}{n}]) = 1 - \frac{1}{n} \uparrow 1, and ∫lim⁡fn=m([0,1))=1\int \lim f_n = m([0, 1)) = 1, so ∫fn↑∫f\int f_n \uparrow \int f holds with equality — the monotone convergence theorem verified on this sequence. No mass escaped and none concentrated at a point. This proof is for this concrete sequence, not a proof of the general monotone convergence theorem.

For the escaping sequence fn=1[n,n+1]f_n = \mathbf{1}_{[n,n+1]} (on [0,∞)[0, \infty)), prove that no integrable function gg dominating every fnf_n exists, and hence the dominated convergence theorem cannot be applied.

∫fn=1  ∀n,lim⁡n→∞fn=0 pointwise,∫lim⁡fn=0\int f_n = 1 \;\forall n, \quad \lim_{n \to \infty} f_n = 0 \text{ pointwise}, \quad \int \lim f_n = 0
g≥fn  ∀n  ⇒  g≥1 on [1,∞)  ⇒  ∫g=∞g \ge f_n \;\forall n \;\Rightarrow\; g \ge 1 \text{ on } [1, \infty) \;\Rightarrow\; \int g = \infty
n=3n = 3: the spike [3,4][3, 4] is outside the [0,2][0, 2] window. ∫f3=1\int f_3 = 1 yet ∫lim⁡fn=0\int \lim f_n = 0 — the mass has already left for infinity.
Show proof
  1. ∫fn=m([n,n+1])=1\int f_n = m([n, n+1]) = 1 for every nn. For fixed x≥0x \ge 0, once n>xn > x we have x∉[n,n+1]x \notin [n, n+1], so fn(x)=0f_n(x) = 0 — the pointwise limit is 00 and ∫lim⁡fn=0≠1=lim⁡∫fn\int \lim f_n = 0 \ne 1 = \lim \int f_n.
  2. Suppose, for contradiction, an integrable g≥0g \ge 0 with ∣fn∣≤g|f_n| \le g for all nn. For each nn, g≥fn=1[n,n+1]g \ge f_n = \mathbf{1}_{[n,n+1]}, so g≥1g \ge 1 on [n,n+1][n, n+1]. Union over all n≥1n \ge 1: g≥1g \ge 1 on [1,∞)=⋃n≥1[n,n+1][1, \infty) = \bigcup_{n \ge 1} [n, n+1] — hence ∫0∞g≥∫1∞1=∞\int_0^\infty g \ge \int_1^\infty 1 = \infty, contradicting integrability.
  3. So no integrable dominating function for this sequence exists — the key hypothesis of the dominated convergence theorem fails, which is exactly why lim⁡∫fn≠∫lim⁡fn\lim \int f_n \ne \int \lim f_n is possible here. When mass escapes to infinity, the integral does not "commute" with the limit. This proof is for this concrete sequence, not a proof of the general dominated convergence theorem.

10. Integration