Measure Theory 11 | Proof of Lebesgue's Dominated Convergence Theorem [dark version]

Q: Why does the proof spend time showing f_n and f are in L1 before tackling convergence?

Because the theorem’s conclusion only makes sense when the integrals are finite. From |f_n| ≤ G a.e. and G ∈ L1, monotonicity gives ∫|f_n| dμ ≤ ∫G dμ < ∞, so each f_n is integrable. Since f_n → f almost everywhere, the same domination implies |f| ≤ G a.e., and the same monotonicity argument yields ∫|f| dμ < ∞. This guarantees all integrals used later are well-defined and finite.

Q: How does the proof turn pointwise convergence into an L1 statement?

It proves the stronger claim ∫|f_n − f| dμ → 0. Using the triangle inequality, |f_n − f| ≤ |f_n| + |f|. With the domination |f_n| ≤ G and |f| ≤ G, this yields |f_n − f| ≤ 2G a.e. The proof then defines h_n = 2G − |f_n − f|, which is nonnegative and measurable, and uses Fatou’s Lemma on h_n to control the integral of |f_n − f|.

Q: What role does Fatou’s Lemma play, and why is h_n constructed as 2G − |f_n − f|?

Fatou’s Lemma provides an inequality relating lim inf of functions to the integral of their lim inf. The construction h_n = 2G − |f_n − f| ensures h_n is nonnegative (a requirement for Fatou) and that h_n(x) → 2G(x) pointwise because |f_n(x) − f(x)| → 0. Applying Fatou to h_n yields an inequality that, after comparing both sides and using the nonnegativity structure, forces the integral of |f_n − f| to converge to 0.

Q: Why can “almost everywhere” inequalities be treated as if they hold everywhere in the proof?

Because integrals ignore changes on sets of measure zero. If |f_n| ≤ G holds only almost everywhere, one can redefine G on the null set so that the inequality holds everywhere without changing ∫G dμ. The proof uses this idea to simplify notation and to avoid carrying “a.e.” through every step.

Q: How does ∫|f_n − f| → 0 imply ∫f_n → ∫f?

By bounding the difference of integrals with the integral of the absolute difference. Using linearity, ∫(f_n − f) dμ = ∫f_n dμ − ∫f dμ. Then the triangle inequality gives |∫(f_n − f) dμ| ≤ ∫|f_n − f| dμ. Since the right-hand side tends to 0, the left-hand side must also tend to 0, yielding ∫f_n dμ → ∫f dμ.

TL;DR

Assuming |f_n| ≤ G a.e. with G integrable guarantees every f_n and the limit f are in L1, so all relevant integrals are finite.

Briefing Cornell Notes

Briefing

Lebesgue’s Dominated Convergence Theorem hinges on a single mechanism: a pointwise limit can be moved inside an integral when every function in the sequence is controlled by one integrable “majorant” G. Concretely, if measurable functions f_n converge to f almost everywhere and there exists an integrable function G such that |f_n(x)| ≤ G(x) for almost every x, then the integrals converge: ∫ f_n dμ → ∫ f dμ. The proof matters because it turns pointwise convergence—often too weak to say anything about integrals—into a reliable statement about areas under curves, which is central in analysis and probability.

The argument starts by checking that the assumptions make the integrals well-defined. Membership in L1 means the integral of |f_n| is finite rather than infinite. Since |f_n| ≤ G and G is integrable, monotonicity of the integral gives ∫ |f_n| dμ ≤ ∫ G dμ < ∞, so each f_n lies in L1. The same domination applies to the limit function f: because f_n → f almost everywhere, the inequality |f(x)| ≤ G(x) holds almost everywhere as well, forcing ∫ |f| dμ < ∞. With integrability secured, the proof can focus entirely on convergence.

The core strengthening shows that the L1 distance between f_n and f goes to zero: ∫ |f_n − f| dμ → 0. The proof uses a pointwise inequality derived from the triangle inequality for absolute values: |f_n − f| ≤ |f_n| + |f|. Since both |f_n| and |f| are dominated by G, one gets |f_n − f| ≤ 2G almost everywhere, and the proof then packages the expression into a nonnegative measurable function h_n(x) = 2G(x) − |f_n(x) − f(x)|. This construction is designed so that h_n(x) increases toward 2G(x) as n grows, because |f_n(x) − f(x)| → 0 pointwise.

Fatou’s Lemma then supplies the key inequality involving lim inf of integrals. Applying it to h_n yields an inequality where the lim inf can be pulled outside the integral, producing a comparison between ∫ 2G dμ and ∫ h_n dμ. After carefully handling the limit superior/smallest possible bounds for the nonnegative term |f_n − f|, the proof forces the inequalities to collapse into equalities. That collapse implies the limit exists and equals 0, giving ∫ |f_n − f| dμ → 0.

Finally, the desired convergence of integrals follows from this L1 convergence. The absolute value of the difference of integrals is bounded by the integral of |f_n − f|, using linearity and the triangle inequality (in integral form). Since ∫ |f_n − f| dμ → 0, the difference ∫ f_n dμ − ∫ f dμ must also go to 0. The dominated convergence theorem emerges as a direct consequence of the domination by an integrable majorant and the Fatou-based strengthening to L1 convergence.

Cornell Notes

Lebesgue’s Dominated Convergence Theorem says that if measurable functions f_n converge to f almost everywhere and there is an integrable majorant G with |f_n| ≤ G a.e., then ∫ f_n dμ → ∫ f dμ. The proof first uses domination and monotonicity to show every f_n and the limit f lie in L1, so the integrals are finite. The key step strengthens the goal: it proves ∫ |f_n − f| dμ → 0. That strengthening is obtained by defining h_n = 2G − |f_n − f| (nonnegative and measurable), applying Fatou’s Lemma to h_n, and forcing inequalities to become equalities. Once L1 convergence is established, the integral convergence follows from bounding |∫(f_n − f)| by ∫|f_n − f|.

Why does the proof spend time showing f_n and f are in L1 before tackling convergence?

Because the theorem’s conclusion only makes sense when the integrals are finite. From |f_n| ≤ G a.e. and G ∈ L1, monotonicity gives ∫|f_n| dμ ≤ ∫G dμ < ∞, so each f_n is integrable. Since f_n → f almost everywhere, the same domination implies |f| ≤ G a.e., and the same monotonicity argument yields ∫|f| dμ < ∞. This guarantees all integrals used later are well-defined and finite.

How does the proof turn pointwise convergence into an L1 statement?

It proves the stronger claim ∫|f_n − f| dμ → 0. Using the triangle inequality, |f_n − f| ≤ |f_n| + |f|. With the domination |f_n| ≤ G and |f| ≤ G, this yields |f_n − f| ≤ 2G a.e. The proof then defines h_n = 2G − |f_n − f|, which is nonnegative and measurable, and uses Fatou’s Lemma on h_n to control the integral of |f_n − f|.

What role does Fatou’s Lemma play, and why is h_n constructed as 2G − |f_n − f|?

Fatou’s Lemma provides an inequality relating lim inf of functions to the integral of their lim inf. The construction h_n = 2G − |f_n − f| ensures h_n is nonnegative (a requirement for Fatou) and that h_n(x) → 2G(x) pointwise because |f_n(x) − f(x)| → 0. Applying Fatou to h_n yields an inequality that, after comparing both sides and using the nonnegativity structure, forces the integral of |f_n − f| to converge to 0.

Why can “almost everywhere” inequalities be treated as if they hold everywhere in the proof?

Because integrals ignore changes on sets of measure zero. If |f_n| ≤ G holds only almost everywhere, one can redefine G on the null set so that the inequality holds everywhere without changing ∫G dμ. The proof uses this idea to simplify notation and to avoid carrying “a.e.” through every step.

How does ∫|f_n − f| → 0 imply ∫f_n → ∫f?

By bounding the difference of integrals with the integral of the absolute difference. Using linearity, ∫(f_n − f) dμ = ∫f_n dμ − ∫f dμ. Then the triangle inequality gives |∫(f_n − f) dμ| ≤ ∫|f_n − f| dμ. Since the right-hand side tends to 0, the left-hand side must also tend to 0, yielding ∫f_n dμ → ∫f dμ.

Review Questions

What integrability and domination conditions are needed to ensure ∫|f_n| dμ and ∫|f| dμ are finite?
Where exactly does the proof use Fatou’s Lemma, and what nonnegative sequence is it applied to?
Why does proving ∫|f_n − f| dμ → 0 automatically give the original dominated convergence conclusion about ∫f_n dμ?

Key Points

1
Assuming |f_n| ≤ G a.e. with G integrable guarantees every f_n and the limit f are in L1, so all relevant integrals are finite.
2
The proof’s main strengthening is L1 convergence: ∫|f_n − f| dμ → 0.
3
Triangle inequality plus domination yields the pointwise bound |f_n − f| ≤ 2G a.e., enabling the construction h_n = 2G − |f_n − f|.
4
Defining h_n makes the sequence nonnegative and measurable, which is exactly what Fatou’s Lemma requires.
5
Fatou’s Lemma applied to h_n produces an inequality involving lim inf that, together with pointwise convergence, forces the integral of |f_n − f| to vanish.
6
Once L1 convergence is known, the dominated convergence conclusion follows from |∫(f_n − f)| ≤ ∫|f_n − f|.
7
“Almost everywhere” domination can be treated as everywhere by redefining functions on null sets without changing integrals.

Highlights

The proof upgrades pointwise convergence to L1 convergence by showing ∫|f_n − f| dμ → 0.

A clever nonnegative auxiliary sequence h_n = 2G − |f_n − f| turns domination into a Fatou-friendly setup.

Fatou’s Lemma plus pointwise convergence forces inequalities to become equalities, leaving no slack.

The final step is a clean inequality: |∫(f_n − f)| ≤ ∫|f_n − f|, so L1 convergence implies integral convergence.

Topics

Dominated Convergence Theorem
Fatou’s Lemma
L1 Convergence
Measurable Functions
Integrable Majorant

Mentioned

L1
a.e.