Limits Under the Integral Sign
Integration is an infinite operation, so the first real question is when it commutes with limits. Three theorems answer — and each one is accompanied by the example that shows what breaks without it.
In this subchapter
5.2.1The question, and a warning
Everything downstream — laws of large numbers (7.2), martingale convergence (8.3), the very definition of conditional expectation (8.1) — needs to move a limit through an expectation: from Xn → X pointwise, conclude E[Xn] → E[X]. Pointwise convergence alone is nowhere near enough:
On ([0,1], ℒ, λ) — or on ℝ with Lebesgue measure — let Xn = n · 1(0, 1/n]: a rectangle of height n and width 1/n. At every fixed ω > 0, eventually 1/n < ω, so Xn(ω) = 0 for all large n: Xn → 0 pointwise. But E[Xn] = n · (1/n) = 1 for every n, so E[Xn] → 1 ≠ 0 = E[lim Xn]. The mass did not vanish; it fled — squeezed into ever-thinner spikes of constant area. Limit and expectation fail to commute, and no amount of pointwise convergence repairs it. Every theorem in this subchapter is a hypothesis under which Example 1's pathology is barred.
5.2.2Monotone Convergence
If 0 ≤ X₁ ≤ X₂ ≤ ⋯ and Xn ↑ X pointwise, then E[Xn] ↑ E[X] (both possibly ∞). In particular lim E[Xn] = E[lim Xn].
By monotonicity of expectation, E[Xn] increases and E[Xn] ≤ E[X] for all n, so lim E[Xn] ≤ E[X]: one inequality is free. For the reverse, price X directly by its banded approximants Yn (5.1 §3). Fix n and ε > 0, and consider the events Cm = {Xm > Yn − ε}. Since Xm ↑ X and X ≥ Yn everywhere, every outcome with X(ω) > Yn(ω) − ε — which is to say every outcome, the inequality being strict by the ε-gap — lands in Cm for all large m; so Cm ↑ Ω. Continuity from below gives P(Cm) → 1, and the good-set split prices E[Xm] ≥ E[Yn] − ‖Yn‖∞P(Cmc) − ε. As before: limm E[Xm] ≥ E[Yn] − ε; let ε → 0, n → ∞, and use Theorem 1 of 5.1 (well-posedness: E[Yn] ↑ E[X]). ∎
Drop nonnegativity, keep monotone: Xn = −1(n,∞) on ℝ with Lebesgue measure. Then Xn ↑ 0 pointwise (each ω eventually has n > ω), but E[Xn] = −∞ for every n while E[0] = 0. Monotone decrease toward the limit from above fails similarly in the infinite-expectation regime: the fix, when you need it, is to apply MCT to the nonnegative parts or to differences known integrable — see the downward version in §5. ∎-adjacent moral: hypotheses in these theorems are not decoration; each names the exact escape route Example 1 and its cousins take.
5.2.3Fatou: the inequality that survives everything
For any sequence Xn ≥ 0 (no monotonicity, no domination):
Recall liminf from 2.2 / 4.1 Theorem 2: the lowest eventual floor — liminf Xn(ω) = limn infm≥n Xm(ω).
Define Zn = infm≥n Xm. Then Zn is nonnegative, Zn ≤ Xn (so E[Zn] ≤ E[Xn] by monotonicity), and Zn ↑ liminf Xn by definition of liminf. MCT gives E[liminf Xn] = lim E[Zn] ≤ liminf E[Xn] — the last step because lim E[Zn] is a limit of numbers each bounded by the corresponding tail infimum. Three lines, and the whole strength is borrowed from MCT. ∎
The moving bump of Example 1: liminf Xn = 0 pointwise, so the left side is 0; the right side is liminf 1 = 1. Fatou's inequality reads 0 ≤ 1 — true, and uselessly loose. Fatou asks for no hypotheses precisely because it promises so little: it is the theorem you cite when you cannot justify anything stronger, and it is always true. A second sharp instance for the other direction of intuition: spikes Xn = n²·1(0,1/n] have liminf Xn = 0 but E[Xn] = n → ∞: the right side can be infinite while the left is zero. Mass can always escape upward; no limit theorem without domination can stop it.
5.2.4Dominated Convergence
Suppose Xn → X pointwise (or even just in probability) and there is a single integrable dominating variable Z ∈ L¹ with |Xn| ≤ Z for all n. Then E|Xn − X| → 0, and in particular E[Xn] → E[X] and E|Xn| → E|X|.
The conclusion E|Xn − X| → 0 is stronger than convergence of expectations, and the route is: apply Fatou to the nonnegative remainders Wn = 2Z − |Xn − X| ≥ 0 (nonnegative since |Xn − X| ≤ |Xn| + |X| ≤ 2Z; note |X| ≤ Z survives the limit). Pointwise, Wn → 2Z. Fatou:
using linearity — legitimate because E[2Z] < ∞ makes every term finite, no ∞ − ∞ (5.1 §5's warning honored). Cancel E[2Z]: limsup E|Xn − X| ≤ 0, so it is 0. Expectations then converge by the triangle |E[Xn] − E[X]| ≤ E|Xn − X|. ∎
(a) Moving bump n·1(0,1/n]: no integrable Z dominates (the needed bound grows with n), DCT cannot be invoked — and indeed its conclusion fails (Example 1). (b) A subtler failure: infinite limit mass on a countable space. Take Ω = ℕ with P({k}) = c/k², c = 6/π² (Basel's sum, and a check that the masses total 1 — 2.1). Let Xn(k) = k·1{k≤n}, so Xn ↑ X(k) = k pointwise. Each partial sum is integrable, and MCT correctly reports E[Xn] = c Σk≤n 1/k → ∞, since E[X] = c Σk 1/k = ∞: the limit is not integrable. DCT's conclusion is unavailable — no integrable Z dominates, and expectations do not converge to a finite limit. The escape route here is different from the bump's: not thinning spikes, but a limit that accumulates genuinely infinite mass. Domination by an integrable Z is precisely the hypothesis that a pointwise limit cannot smuggle in new mass.
The classical payoff: φ(t) = E[cos(tX)] (the characteristic function — 7.3 will make it central). Is φ differentiable, with φ'(t) = E[−X sin(tX)]? Form the difference quotient (φ(t+h) − φ(t))/h = E[ X·(cos((t+h)X) − cos(tX))/(hX) ] — the bracket is bounded by 1 in absolute value (a mean-value bound: the difference quotient of cosine against its own argument). If E|X| < ∞, the variable |X| dominates the integrands, DCT passes the limit in h through E, and φ'(t) = −E[X sin(tX)]. If E|X| = ∞ — Cauchy variables, 5.5 — the argument collapses at the domination step, and indeed the Cauchy characteristic function e−|t| is not differentiable at 0: the failure of a derivative is the visible scar of a failed DCT hypothesis.
5.2.5The theorems in the wild
How many of the events A₁, A₂, … occur? The variable N = Σn 1An — an infinite sum of indicators, well-defined as the pointwise limit of partial sums (measurable by 4.1 Theorem 2). If the partial sums are bounded by an integrable Z, DCT gives E[N] = Σ P(An); in general (nonnegative), MCT gives the same identity in [0,∞]: the expected count is the sum of probabilities. So "Σ P(An) < ∞ implies only finitely many An occur" (Borel–Cantelli I, 6.3) is just: E[N] finite forces N finite a.s. — a first taste of how limit theorems collapse chapters of machinery into lines.
5.2.6What cannot be rescued
There is no theorem "pointwise convergence implies expectation convergence", because Example 1 exists. The three theorems are the complete repair kit under extra hypotheses, and their relationships are worth memorizing as a flow: MCT (monotone, nonnegative) → Fatou (arbitrary, nonnegative, inequality only) → DCT (dominated, equality). Each is proved from the previous; each weakens one hypothesis and weakens the conclusion accordingly. When a derivation downstream needs to pass a limit through E, the discipline is: name the theorem, name the hypothesis, check the hypothesis on the page — the same brittleness-flagging habit the course has trained since 1.1. A proof that silently swaps a limit and an expectation is not a proof; it is a wish.
Limits pass through expectations exactly when mass cannot escape — monotone, bounded below, or dominated. Every hypothesis is a fence against Example 1.
The integral is built and safe under limits. The next subchapter steps back and compares it with the integral a quant already owns — Riemann's — proving that Lebesgue extends it, exhibiting the function that separates them (the indicator of the rationals), and settling when the two agree so that everyday calculus remains trustworthy.