Phase 6 - Lesson 6.4
Convergence Theorems: Monotone Convergence, Fatou, Dominated Convergence
The three permits for swapping a limit and an integral - the reason the Lebesgue integral was worth building.
Builds on: The integral of 6.3 becomes powerful only with these interchange theorems.
Leads to: Dominated convergence justifies differentiating under expectation throughout Phases 7–15.
Leads to: Dominated convergence justifies differentiating under expectation throughout Phases 7–15.
Learning Objectives
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- State the monotone convergence, Fatou, and dominated convergence theorems.
- Give a concrete example where limit and integral do NOT commute.
- Apply dominated convergence to justify swapping \(\lim\) and \(\int\) with a dominating function.
- Explain how Fatou provides a one-sided inequality without domination.
- Interpret these theorems as the analytic backbone of expectation limits.
Key Vocabulary
- Monotone convergence (MCT)
- If \(0\le f_n\uparrow f\) then \(\int f_n\to\int f\); limits pass through for increasing nonnegative sequences.
- Fatou's lemma
- For \(f_n\ge0\), \(\int\liminf f_n\le\liminf\int f_n\); a one-sided inequality, always valid.
- Dominated convergence (DCT)
- If \(f_n\to f\) a.e. and \(|f_n|\le g\in L^1\), then \(\int f_n\to\int f\).
- Dominating function
- An integrable \(g\) bounding all \(|f_n|\); the hypothesis that licenses DCT.
- Escape of mass
- Failure of interchange when probability/mass drifts to infinity or spikes, e.g. \(n\mathbf{1}_{(0,1/n)}\).
- Uniform integrability
- A sharpening of domination controlling tails uniformly; the general condition behind \(L^1\) convergence.
Intuition & Motivation
Intuition
The whole point of Lebesgue’s theory is that limits and integrals usually commute - but not always. The three theorems are graded permits. MCT: if the functions only rise, mass can never sneak away, so \(\lim\int=\int\lim\) for free. Fatou: with no control at all you still get a one-sided guarantee - the integral of the limit can only be as small as, or smaller than, the liminf of the integrals. DCT: if a single integrable \(g\) caps every \(|f_n|\), mass cannot escape and full equality returns. The cautionary tale \(f_n=n\mathbf{1}_{(0,1/n)}\) has no such cap, and interchange fails.
The three theorems
Theorem - Monotone Convergence
If \(f_n\) are measurable with \(0\le f_1\le f_2\le\cdots\) and \(f_n\to f\) pointwise, then \(\lim_n\int f_n\,d\mu=\int f\,d\mu\) (both sides possibly \(+\infty\)).
Lemma - Fatou
For any measurable \(f_n\ge0\), \(\int\big(\liminf_n f_n\big)\,d\mu\le\liminf_n\int f_n\,d\mu\). No monotonicity or domination is required; only the inequality direction is guaranteed.
Theorem - Dominated Convergence
If \(f_n\to f\) almost everywhere and there is \(g\in L^1\) with \(|f_n|\le g\) for all \(n\), then \(f\in L^1\) and \(\lim_n\int f_n\,d\mu=\int f\,d\mu\); moreover \(\int|f_n-f|\to0\).
Proof
DCT from Fatou: apply Fatou to \(g+f_n\ge0\) and to \(g-f_n\ge0\). The first gives \(\int f\le\liminf\int f_n\); the second gives \(\int f\ge\limsup\int f_n\). Squeezing yields \(\int f_n\to\int f\). ∎
Worked Example - The counterexample where interchange fails
1
Let \(f_n(x)=n\,\mathbf{1}_{(0,1/n)}(x)\) on \([0,1]\): a spike of height \(n\) and width \(1/n\).
2
Integral: \(\int f_n\,d\mu=n\cdot\tfrac1n=1\) for every \(n\), so \(\lim_n\int f_n=1\).
3
Pointwise limit: for any fixed \(x\gt 0\), eventually \(1/n\lt x\) so \(f_n(x)=0\); also \(f_n(0)=0\). Hence \(f_n\to0\) everywhere, and \(\int\lim f_n=0\).
4
\(1=\lim\int f_n\ne\int\lim f_n=0\): the mass ‘escaped’ into an ever-thinner, taller spike. No integrable \(g\ge n\mathbf{1}_{(0,1/n)}\) for all \(n\) exists, so DCT does not apply - and Fatou’s \(0\le1\) holds strictly.
Interactive: watch the limit and integral disagree
Compute the constant integral of the escaping-spike family and its pointwise limit at a fixed point - the two numbers famously differ.
Common Mistakes to Avoid
- Swapping \(\lim\) and \(\int\) with no justification - always cite MCT, Fatou, or DCT.
- Applying DCT without producing an explicit integrable dominator \(g\).
- Expecting Fatou to give equality - it is only the inequality \(\le\), and can be strict.
- Using MCT on a decreasing or sign-changing sequence - it requires \(0\le f_n\uparrow\).
Quant Practitioner Tips
- DCT is the everyday workhorse: find a dominator, and limits pass through integrals (and expectations).
- To differentiate under the integral sign, dominate the difference quotients and apply DCT.
- When you only need a bound (not equality) and have no dominator, reach for Fatou.
- ‘Escaping mass’ (tall thin spikes or drift to \(\infty\)) is the signal that a dominator cannot exist.
Knowledge Check
Q1 Medium
For \(f_n=n\mathbf{1}_{(0,1/n)}\) on \([0,1]\), which statement is correct?
Q2 Easy
Which theorem gives a one-sided inequality valid with NO domination or monotonicity hypothesis?
Q3 Medium
To apply the dominated convergence theorem you must exhibit:
Practical Exercise
Use dominated convergence to evaluate \(\lim_{n\to\infty}\int_0^1\frac{n\,x}{1+n^2x^2}\cdot\frac{1}{n}\,dx\), i.e. \(\lim_n\int_0^1\frac{x}{1+n^2x^2}\,dx\), by finding a dominator and the pointwise limit.
▶ Show full solution
Let \(f_n(x)=\dfrac{x}{1+n^2x^2}\) on \([0,1]\). For fixed \(x\gt 0\), \(f_n(x)\to0\) as \(n\to\infty\); and \(f_n(0)=0\). So \(f_n\to0\) pointwise.
Dominator: since \(1+n^2x^2\ge1\) and \(0\le x\le1\), we have \(0\le f_n(x)\le x\le1=:g(x)\), and \(g\in L^1[0,1]\).
By DCT, \(\lim_n\int_0^1 f_n\,dx=\int_0^1 0\,dx=0\).
After the reveal, answer for yourself: Where would this argument break if the domain were \([0,\infty)\) instead of \([0,1]\)?
Lesson Summary
The three convergence theorems say when \(\lim\int=\int\lim\). MCT handles increasing nonnegative sequences; Fatou gives the always-valid one-sided \(\int\liminf\le\liminf\int\); DCT restores equality given an integrable dominator. The spike \(n\mathbf{1}_{(0,1/n)}\) shows interchange can genuinely fail when mass escapes.
Retrieval Practice
Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.
▶ Show retrieval prompts & answers
Q: State dominated convergence and its key hypothesis.
A: If \(f_n\to f\) a.e. and \(|f_n|\le g\in L^1\), then \(\int f_n\to\int f\) and \(\int|f_n-f|\to0\); the hypothesis is a single integrable dominator \(g\).
A: If \(f_n\to f\) a.e. and \(|f_n|\le g\in L^1\), then \(\int f_n\to\int f\) and \(\int|f_n-f|\to0\); the hypothesis is a single integrable dominator \(g\).
Q: Give an example where limit and integral do not commute.
A: \(f_n=n\mathbf{1}_{(0,1/n)}\) on \([0,1]\): \(\int f_n=1\) for all \(n\) but \(f_n\to0\) pointwise, so \(\lim\int=1\ne0=\int\lim\).
A: \(f_n=n\mathbf{1}_{(0,1/n)}\) on \([0,1]\): \(\int f_n=1\) for all \(n\) but \(f_n\to0\) pointwise, so \(\lim\int=1\ne0=\int\lim\).
Completion Checklist
- I can explain the core ideas in my own words
- I worked the derivations/examples by hand
- I completed the interactive workbench(es)
- I passed the knowledge check
Confidence / mastery rating
Personal notes
Source References
This lesson synthesizes and paraphrases concepts from the sources below. No copyrighted text, problem sets, or solutions are reproduced. Return to the originals for full depth.
- Measure, Integration & Real Analysis (Sheldon Axler, GTM 282) current - Ch. 3 - Ch. 3: monotone convergence, Fatou, dominated convergence.
- Probability: Theory and Examples (Rick Durrett, 5th ed.) current - Ch. 1 - Ch. 1: convergence theorems in the probability setting.