Lebesgue Convergence Theorems

Three theorems license swapping limits and integrals. Examples with escaping mass show why each hypothesis is needed.

Swapping the order requires that no mass leaks away. A dominating function overhead keeps things safe. Without conditions, area can vanish in the limit.

For example, pour πŸ’§ water from cup to cup with no spill and the total stays the same. One drop escaping sideways shrinks the total that arrives.

fn↑fβ‡’βˆ«fnβ†‘βˆ«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
∣fnβˆ£β‰€g,Β fnβ†’fΒ a.e.Β β‡’βˆ«fnβ†’βˆ«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=nβ‹…1n=1βˆ€n\int f_n = n \cdot \frac{1}{n} = 1 \quad \forall n
∫lim inf⁑nβ†’βˆžfn=0<1=lim inf⁑nβ†’βˆžβˆ«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 β€” 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=nβ‹…m([0,1/n])=nβ‹…1n=1\int f_n = n \cdot m([0, 1/n]) = n \cdot \frac{1}{n} = 1 β€” the same for every nn. Hence lim inf⁑nβ†’βˆžβˆ«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 β€” so lim inf⁑nfn(x)=0\liminf_n f_n(x) = 0.

    At x=0x = 0, fn(0)=nβ†’βˆžf_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 β€” 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[0,1βˆ’1/n]f_n = \mathbf{1}_{[0,1-1/n]} increases to f=1[0,1)f = \mathbf{1}_{[0,1)}, and prove ∫fnβ†‘βˆ«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=1βˆ’1n↑1=∫lim⁑fn\int f_n = 1 - \frac{1}{n} \uparrow 1 = \int \lim f_n
n=10n = 10: ∫f10=1βˆ’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 1βˆ’1n1 - \frac{1}{n} increases, [0,1βˆ’1n]βŠ‚[0,1βˆ’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} β€” a monotone increasing sequence of nonnegative measurable functions.

  2. Pointwise limit: if x<1x < 1, eventually 1βˆ’1n>x1 - \frac{1}{n} > x, so fn(x)=1f_n(x) = 1 and the limit is 11.

    If xβ‰₯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([0,1βˆ’1n])=1βˆ’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 ∫fnβ†‘βˆ«f\int f_n \uparrow \int f holds with equality β€” 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=1β€…β€Šβˆ€n,lim⁑nβ†’βˆžfn=0Β 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
gβ‰₯fnβ€…β€Šβˆ€nβ€…β€Šβ‡’β€…β€Šgβ‰₯1Β onΒ [1,∞)β€…β€Šβ‡’β€…β€Šβˆ«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 β€” 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 xβ‰₯0x \ge 0, once n>xn > x we have xβˆ‰[n,n+1]x \notin [n, n+1], so fn(x)=0f_n(x) = 0 β€” the pointwise limit is 00 and ∫lim⁑fn=0β‰ 1=lim⁑∫fn\int \lim f_n = 0 \ne 1 = \lim \int f_n.

  2. Suppose, for contradiction, an integrable gβ‰₯0g \ge 0 with ∣fnβˆ£β‰€g|f_n| \le g for all nn.

    For each nn, gβ‰₯fn=1[n,n+1]g \ge f_n = \mathbf{1}_{[n,n+1]}, so gβ‰₯1g \ge 1 on [n,n+1][n, n+1].

    Union over all nβ‰₯1n \ge 1: gβ‰₯1g \ge 1 on [1,∞)=⋃nβ‰₯1[n,n+1][1, \infty) = \bigcup_{n \ge 1} [n, n+1] β€” hence ∫0∞gβ‰₯∫1∞1=∞\int_0^\infty g \ge \int_1^\infty 1 = \infty, contradicting integrability.

  3. So no integrable dominating function for this sequence exists β€” the key hypothesis of the dominated convergence theorem fails, which is exactly why lim⁑∫fnβ‰ βˆ«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.

11. Integration