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Measure Theory 9 | Fatou's Lemma [dark version] thumbnail

Measure Theory 9 | Fatou's Lemma [dark version]

4 min read

Based on The Bright Side of Mathematics's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.

TL;DR

Fatou’s Lemma applies to any sequence of non-negative measurable functions on a measure space.

Briefing

Fatou’s Lemma gives a one-sided way to move “liminf” through an integral for non-negative measurable functions. Instead of asking whether an integral of a pointwise limit equals the limit of integrals, it focuses on the limit inferior: for a sequence (f_n) of non-negative measurable functions on a measure space (X, Σ, μ), Fatou’s Lemma guarantees

∫_X liminf_{n→∞} f_n dμ ≤ liminf_{n→∞} ∫_X f_n dμ.

That inequality is weaker than a full convergence theorem, but it is powerful because it requires almost nothing beyond measurability and non-negativity. The lemma’s generality makes it a workhorse for later results—most notably Fatou’s Lemma is positioned as a stepping stone toward the more familiar “reverse” convergence statement known as Fatou’s lemma’s companion, which leads into the next theorem in the series.

A key part of the argument is understanding what liminf_{n→∞} f_n means as a function. For each x in X, liminf_{n→∞} f_n(x) is defined using the tail behavior of the sequence of numbers f_k(x):

liminf_{n→∞} f_n(x) = lim_{n→∞} (inf_{k≥n} f_k(x)).

Because the infimum over k≥n can be 0 or potentially grow without bound, the construction naturally lives in the extended non-negative reals (allowing ∞). This choice matters for the statement’s strength: including ∞ from the start avoids edge cases where the liminf might otherwise be undefined.

The proof also leans on measurability and monotonicity. The function g(x) := liminf f_n(x) is measurable because it is built from measurable functions using operations that preserve measurability: taking infima over tails and then taking limits. To make the monotone convergence theorem usable, the proof defines auxiliary functions g_n(x) := inf_{k≥n} f_k(x). These g_n form an increasing sequence: shifting the cutoff n to the right can only increase the infimum, so g_1 ≤ g_2 ≤ g_3 ≤ … .

With that setup, the monotone convergence theorem turns the limit of integrals into an equality:

∫_X lim_{n→∞} g_n dμ = lim_{n→∞} ∫_X g_n dμ.

Since g_n(x) is always ≤ f_n(x) (because g_n takes the infimum over a set that includes f_n), one gets g_n ≤ f_n pointwise, and therefore ∫ g_n dμ ≤ ∫ f_n dμ by monotonicity of the integral. Combining these steps yields the desired inequality with liminf on both sides.

In short: Fatou’s Lemma turns tail infima into a measurable increasing sequence, uses monotone convergence to pass a limit through the integral, and then compares those tail infima to the original functions. The result is a robust inequality that often replaces stronger convergence assumptions when only non-negative measurability is available.

Cornell Notes

Fatou’s Lemma provides a one-sided inequality for non-negative measurable functions: the integral of the pointwise limit inferior is at most the limit inferior of the integrals. For a sequence (f_n), it asserts ∫ liminf f_n dμ ≤ liminf ∫ f_n dμ. The proof rewrites liminf as a limit of tail infima: liminf f_n(x) = lim_{n→∞} inf_{k≥n} f_k(x). Defining g_n(x) = inf_{k≥n} f_k(x) produces an increasing sequence of measurable functions, letting the monotone convergence theorem move the limit through the integral. Since g_n ≤ f_n pointwise, monotonicity of the integral yields the final inequality.

How does Fatou’s Lemma differ from a standard convergence theorem involving limits?

It does not compare ∫ lim f_n dμ with lim ∫ f_n dμ. Instead, it uses liminf. The lemma guarantees an inequality: ∫_X liminf_{n→∞} f_n dμ ≤ liminf_{n→∞} ∫_X f_n dμ. That one-sided control is weaker than full convergence equality, but it works under very mild assumptions: only non-negative measurability is required.

What is the functional meaning of liminf_{n→∞} f_n(x)?

For each x, liminf is defined via tail infima of the real (or extended real) numbers f_k(x): liminf_{n→∞} f_n(x) = lim_{n→∞} (inf_{k≥n} f_k(x)). This construction naturally allows the value ∞, so the codomain is taken as non-negative extended reals to keep the definition valid in all cases.

Why does defining g_n(x) = inf_{k≥n} f_k(x} help the proof?

Because the sequence (g_n) is monotone increasing: g_1(x) ≤ g_2(x) ≤ g_3(x) ≤ … . Moving the cutoff n to the right shrinks the set {k≥n}, so the infimum over a smaller tail cannot decrease. That monotonicity is exactly what the monotone convergence theorem needs.

Where does measurability enter the argument?

The function g_n(x) = inf_{k≥n} f_k(x) is measurable because it is built from measurable functions using infima over tails. Then g(x) = lim_{n→∞} g_n(x) = liminf f_n(x) is measurable as a limit of measurable functions. This ensures the integrals in the proof are well-defined.

How does the proof connect g_n to f_n to produce the inequality?

By definition, g_n(x) is the infimum over all k≥n, so the particular term f_n(x) is among those values. Therefore g_n(x) ≤ f_n(x) for every x. Pointwise inequality implies integral inequality: ∫ g_n dμ ≤ ∫ f_n dμ. Taking liminf on the right side leads to Fatou’s inequality.

Review Questions

  1. State Fatou’s Lemma precisely, including the direction of the inequality and the role of liminf.
  2. Explain how liminf_{n→∞} f_n(x) can be written using inf_{k≥n} f_k(x).
  3. In the proof, why is the sequence g_n(x) = inf_{k≥n} f_k(x) monotone increasing?

Key Points

  1. 1

    Fatou’s Lemma applies to any sequence of non-negative measurable functions on a measure space.

  2. 2

    The lemma controls liminf through integration: ∫ liminf f_n dμ ≤ liminf ∫ f_n dμ.

  3. 3

    The pointwise liminf can be rewritten as lim_{n→∞} inf_{k≥n} f_k(x).

  4. 4

    Defining g_n(x) = inf_{k≥n} f_k(x) yields a measurable, monotonically increasing sequence.

  5. 5

    Monotone convergence theorem converts ∫ lim g_n dμ into lim ∫ g_n dμ.

  6. 6

    Because g_n ≤ f_n pointwise, monotonicity of the integral gives ∫ g_n dμ ≤ ∫ f_n dμ.

  7. 7

    The result is one-sided and weaker than a full convergence theorem, but it needs only non-negativity and measurability.

Highlights

Fatou’s Lemma replaces “limit” with “limit inferior,” guaranteeing ∫ liminf f_n dμ ≤ liminf ∫ f_n dμ.
Writing liminf f_n(x) as lim_{n→∞} inf_{k≥n} f_k(x} turns the problem into one about tail infima.
The monotone convergence theorem drives the proof once g_n(x) = inf_{k≥n} f_k(x) is shown to increase with n.
The inequality ultimately comes from the simple bound inf_{k≥n} f_k(x) ≤ f_n(x), which transfers to integrals.

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