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
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