The convergence theorems of Lebesgue's theory are a hierarchy of bargains. The Monotone Convergence Theorem lets you swap a limit and an integral — but only if the functions climb monotonically. Dominated Convergence lets you swap too — but only if a single integrable function dominates the whole sequence. Both hand you an equality, and both demand a hypothesis you must go out and verify.

Fatou's Lemma is the member of the family that asks for almost nothing. No monotonicity. No dominating function. No convergence at all — the sequence need not even have a limit. The single price of admission is non-negativity: f_n \ge 0. In exchange for demanding so little, it pays you back with less — not an equality but an inequality. It is the safety net you can always deploy, the one convergence result that never asks whether its hypotheses hold, because they almost always do. This page teaches that one idea: in the limit, mass can only be lost, never gained.

The statement: the integral of the liminf is at most the liminf of the integrals

Let (f_n) be any sequence of non-negative measurable functions on a measure space (X, \mathcal{M}, \mu). Then

Read it slowly, because the whole lesson is in the order of operations. On the left we first take the \liminf of the functions — a pointwise construction that builds a new function g = \liminf_n f_n — and only then integrate it. On the right we first integrate each f_n to get a sequence of numbers \int f_n, and take the \liminf of those. The claim is that the first quantity can never exceed the second.

Think of \int f_n as the "mass" carried by f_n. Fatou says that when you pass to the limit, some of that mass may leak away — the limiting function can hold less than the sequence carried — but no mass is ever conjured from nowhere. The limit is stingy. It keeps at most what the sequence brought, and sometimes less.

The proof: two lines, once you have Monotone Convergence

Fatou's Lemma is not proved from scratch — it is a short corollary of the Monotone Convergence Theorem (MCT). The trick is to unfold the definition of \liminf into an increasing sequence, exactly the shape MCT wants.

Step 1 — build the running infima. For each n set

Each g_n is non-negative and measurable (a countable infimum of measurable functions). As n grows we take the infimum over a smaller tail, so g_n can only rise: g_1 \le g_2 \le g_3 \le \cdots. And by the very definition of the limit inferior, this increasing sequence climbs to

Step 2 — compare each g_n to the tail. Since g_n = \inf_{k\ge n} f_k \le f_k for every k \ge n, monotonicity of the integral gives \int g_n \le \int f_k for all such k. Taking the infimum over the tail on the right,

Step 3 — apply MCT and let n \to \infty. Because 0 \le g_n \uparrow \liminf_n f_n, the Monotone Convergence Theorem lets us pull the limit through the integral on the left. On the right, the infima \inf_{k\ge n}\int f_k increase to \liminf_n \int f_n by definition. Chaining it all together:

Two genuine ideas — replace f_n by the running infimum g_n, then feed the monotone g_n to MCT — and the lemma falls out. The inequality enters at exactly one place: Step 2, where g_n throws away everything the tail had above its floor. That discarded surplus is precisely the mass that can escape.

Why the inequality is real: mass that escapes to infinity

Is the \le ever strict? Emphatically yes — and understanding how is understanding the lemma. There are two classic escape artists, and both start from the same trick: keep the area at exactly 1 while pushing the function so that at every fixed point it eventually vanishes.

The tall spike. On \mathbb{R} with Lebesgue measure, take

The spike grows taller and thinner, always enclosing area 1. But fix any point x > 0: once n > 1/x the interval (0, 1/n) no longer contains x, so f_n(x) = 0 from then on. Hence f_n \to 0 pointwise and \liminf_n f_n = 0. Now compare the two sides of Fatou:

So Fatou reads 0 \le 1 — strict. The whole unit of mass has slipped out through the shrinking base; the pointwise limit never sees it. This is why the lemma cannot be upgraded to an equality in general: here the gap is the entire mass.

The sliding bump. Same phenomenon, escaping the other way — to the right instead of upward:

A unit block of area 1 marching off toward +\infty. Fix any x: once n > x the block has moved past it, so again f_n(x) \to 0, \liminf_n f_n = 0, and \int \liminf_n f_n = 0 < 1 = \liminf_n \int f_n. The mass has "escaped to infinity" — carried off to where no fixed point can follow it. Drag the slider below to watch either escape happen.

What Fatou is good for

A one-directional inequality with no hypotheses sounds weak. In practice it is one of the most-used tools in analysis, precisely because you can reach for it unconditionally. Three staple uses:

• It bounds the limit from the sequence. If you know \int f_n \le M for all n and f_n \to f pointwise, Fatou instantly gives \int f = \int \liminf f_n \le \liminf \int f_n \le M. So the limit function is integrable, with its integral controlled by the uniform bound — no monotonicity or domination needed to get the bound.
• It proves the Dominated Convergence Theorem. If |f_n| \le g with g integrable and f_n \to f, apply Fatou to the two non-negative sequences g + f_n \ge 0 and g - f_n \ge 0. The first gives \int f \le \liminf \int f_n; the second gives -\int f \le \liminf(-\int f_n) = -\limsup \int f_n, i.e. \limsup \int f_n \le \int f. Squeeze the two and \int f_n \to \int f. Fatou is the engine inside DCT.
• It guarantees finiteness / Fatou for probability. In probability (\mu a probability measure, f_n = X_n random variables \ge 0), Fatou reads \mathbb{E}[\liminf X_n] \le \liminf \mathbb{E}[X_n] — the expected value of the limiting outcome cannot exceed the limiting expectation. It is the workhorse behind many almost-sure convergence arguments.