Fatou's lemma

Theorem

Fatou's lemma

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Statement

Let \( (X, \mathcal{M}, \mu) \) be a measure space and let \( (f_n)_{n=1}^{\infty} \) be a sequence of \( \mathcal{M} \)-measurable functions \( f_n : X \to [0,\infty] \) (i.e. each \( f_n \) is measurable with values in the extended non-negative reals). Then \[ \int_X \left( \liminf_{n\to\infty} f_n \right) \, d\mu \ \le\ \liminf_{n\to\infty} \int_X f_n \, d\mu. \] In particular both sides are well-defined elements of \( [0,\infty] \) (the integrand \( \liminf_n f_n \) is measurable and non-negative, so its integral exists in \( [0,\infty] \) even if some or all of the \( \int_X f_n\,d\mu \) equal \( +\infty \)).

Why it matters

Fatou's lemma is the workhorse inequality that makes measure-theoretic integration robust under limits without any hypothesis of monotonicity or domination. It asks only for non-negativity and measurability — hypotheses that are essentially always available — and in exchange delivers a one-directional but universally valid comparison between the limit of integrals and the integral of the limit. Every other convergence theorem in the toolkit (Monotone Convergence, Dominated Convergence, Vitali) is either a special case of Fatou or is proved by applying Fatou twice, once to \( f_n \) and once to \( -f_n \) (or to a dominating function minus \( f_n \)).

Its practical value is that it converts a statement about a limit — often hard to control termwise — into a statement about a \( \liminf \) of a sequence of numbers, which is a purely real-variable object. This is exactly the mechanism used throughout probability (Fatou for expectations of non-negative random variables), PDE theory (lower semicontinuity of energy functionals), and functional analysis (weak lower semicontinuity of norms).

Hypotheses

Each \(f_n\) is measurable.Without measurability \( \int_X f_n\,d\mu \) is not even defined, and neither is \( \int_X \liminf_n f_n\,d\mu \) unless \( \liminf_n f_n \) happens to be measurable; the statement is meaningless outside the measurable category. (Note: a pointwise \(\liminf\) of measurable functions is automatically measurable — this is a lemma used inside the proof, not an extra hypothesis.)

Each \(f_n\) is non-negative (valued in \([0,\infty]\)).Drop this: let \( f_n = -\mathbf{1}_{[n,\infty)} \) on \( (\mathbb{R}, \mathcal{L}, \text{Leb}) \). Then \( f_n \to 0 \) pointwise so \( \liminf_n f_n = 0 \) and \( \int 0\,d\mu = 0 \), but \( \int_{\mathbb{R}} f_n\,d\mu = -\infty \) for every \( n \), so \( \liminf_n \int f_n\,d\mu = -\infty \). Now \( 0 \le -\infty \) fails: the inequality reverses. Non-negativity (or a uniform lower bound integrable function, as in the extended version) is what rules this out.

\(\mu\) is a (countably additive, non-negative) measure on the \(\sigma\)-algebra \(\mathcal{M}\).The proof leans on countable additivity through the Monotone Convergence Theorem, which itself uses countable additivity to show that integrals of an increasing sequence of simple functions converge to the integral of the limit. If \(\mu\) were only finitely additive, Monotone Convergence can fail, and so can Fatou: e.g. on \( \mathbb{N} \) with the finitely-additive "density" measure that assigns \( 1 \) to every cofinite set and \( 0 \) to every finite set, one can construct \( f_n = \mathbf{1}_{\{n\}} \) with \( \liminf_n f_n = 0 \) but individual integrals not behaving as Fatou would require under the naive extension of the argument.

Proof

g_n := \inf_{k \ge n} f_k, \qquad n = 1,2,3,\ldots

Definition of the auxiliary sequence used to unpack \(\liminf\). Each \(g_n\) is measurable because a pointwise infimum of countably many measurable \([0,\infty]\)-valued functions is measurable: for every \(a \in \mathbb{R}\), \( \{x : g_n(x) \lt a\} = \bigcup_{k\ge n}\{x: f_k(x) \lt a\} \), a countable union of measurable sets, hence measurable. A

0 \le g_1 \le g_2 \le g_3 \le \cdots, \qquad g_n \uparrow \liminf_{k\to\infty} f_k \text{ pointwise on } X

Monotonicity: \( g_{n+1} = \inf_{k\ge n+1} f_k \ge \inf_{k \ge n} f_k = g_n \) because the infimum is taken over a smaller index set. Non-negativity of each \(g_n\) follows since every \(f_k \ge 0\). By definition, \( \liminf_{k\to\infty} f_k(x) := \lim_{n\to\infty} \inf_{k\ge n} f_k(x) = \lim_{n\to\infty} g_n(x) \), and this limit exists in \([0,\infty]\) for every \(x\) because \((g_n(x))_n\) is a non-decreasing sequence in \([0,\infty]\) (a monotone sequence in the extended reals always converges, possibly to \(+\infty\)). A

g := \liminf_{k\to\infty} f_k \text{ is measurable}

\(g\) is the pointwise limit of the measurable functions \(g_n\) (Step 1), and a pointwise limit of measurable functions is measurable: \( \{x : g(x) \lt a\} = \bigcup_{m=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n\ge N}\{x : g_n(x) \lt a - 1/m\} \) is a countable combination of measurable sets (this is the standard "measurability is preserved under countable sup/inf/lim" lemma). This closes the gap flagged in Hypothesis 1. A

\int_X g_n \, d\mu \ \longrightarrow\ \int_X g \, d\mu \qquad (n \to \infty)

Apply the Monotone Convergence Theorem (MCT) to the sequence \((g_n)\): it is a non-decreasing sequence of measurable functions \(X \to [0,\infty]\) with pointwise limit \(g\) (Steps 1–3), so MCT gives \( \lim_n \int_X g_n\,d\mu = \int_X \lim_n g_n\,d\mu = \int_X g\,d\mu \), and in particular the limit on the left exists in \([0,\infty]\). This is the one substantive external input the proof uses. B

\forall n,\ \forall k \ge n:\quad g_n \le f_k \quad\Longrightarrow\quad \int_X g_n\,d\mu \ \le\ \int_X f_k \, d\mu

\(g_n = \inf_{k\ge n} f_k \le f_k\) for every \(k \ge n\), directly from the definition of infimum. Monotonicity of the integral for non-negative measurable functions (if \(0\le u \le v\) pointwise then \(\int u\,d\mu \le \int v\,d\mu\), itself immediate from the definition of the integral as a supremum over simple functions below the integrand, since every simple function \(\le u\) is also \(\le v\)) then gives the displayed inequality. A

\int_X g_n \, d\mu \ \le\ \inf_{k \ge n} \int_X f_k \, d\mu \qquad \text{for every } n

Step 5 holds for every \(k \ge n\) with the left side fixed, so \( \int_X g_n\,d\mu \) is a lower bound for the set \( \{\int_X f_k\,d\mu : k \ge n\} \); the infimum is the greatest lower bound, hence \(\int_X g_n\,d\mu \le \inf_{k\ge n}\int_X f_k\,d\mu \). B

\int_X g \, d\mu \;=\; \lim_{n\to\infty}\int_X g_n\,d\mu \;\le\; \lim_{n\to\infty}\ \inf_{k\ge n}\int_X f_k\,d\mu \;=\; \liminf_{n\to\infty}\int_X f_n\,d\mu

The equality on the left is Step 4. The middle inequality follows by taking \(n \to \infty\) on both sides of Step 6: the right-hand sequence \( a_n := \inf_{k\ge n}\int_X f_k\,d\mu \) is non-decreasing in \(n\) (same argument as Step 2, applied to the real/extended-real sequence \(\int_X f_k\,d\mu\)) and dominates the left sequence termwise, so passing to the limit preserves the inequality (limits of monotone sequences in \([0,\infty]\) preserve termwise \(\le\)). The last equality is the definition of \(\liminf\) applied to the sequence \(\bigl(\int_X f_n\,d\mu\bigr)_n\). C

\int_X \liminf_{n\to\infty} f_n \, d\mu \ \le\ \liminf_{n\to\infty} \int_X f_n \, d\mu

Substitute \(g = \liminf_n f_n\) (Step 2) into Step 7. This is exactly the claimed inequality, completing the proof. \(\blacksquare\) A

Result

\int_X \liminf_{n\to\infty} f_n \, d\mu \ \le\ \liminf_{n\to\infty} \int_X f_n \, d\mu \qquad (f_n \ge 0 \text{ measurable})

Reading. If you have a sequence of non-negative measurable functions, the "mass" of the pointwise-liminf function can only be less than or equal to what the liminf of the individual masses suggests — mass can disappear in the limit (escape to infinity, oscillate away, concentrate on a shrinking set) but it can never spontaneously appear. Fatou is the "no free lunch" half of the interchange of limit and integral.

Scope. Holds on every measure space (no finiteness of \(\mu\) required, \(\sigma\)-finiteness not needed) for every sequence of \([0,\infty]\)-valued measurable functions; no pointwise convergence of \((f_n)\) is assumed — only \(\liminf\) is used, which always exists. If the \(f_n\) do converge pointwise \(\mu\)-a.e. to some \(f\), the conclusion reads \( \int f\,d\mu \le \liminf_n \int f_n\,d\mu \). The reverse inequality is false in general (see Fails without).

Corollaries & converses

Monotone Convergence Theorem is recovered: if additionally \(f_n \uparrow f\) pointwise, Fatou gives \(\int f \le \liminf \int f_n\), while monotonicity of the integral gives \(\int f_n \le \int f\) for all \(n\) hence \(\limsup \int f_n \le \int f\); combined, \(\int f_n \to \int f\).
Reverse Fatou: if \(f_n \le h\) pointwise for all \(n\) with \(\int h\,d\mu \lt \infty\), applying Fatou to \(h - f_n \ge 0\) gives \(\limsup_n \int f_n\,d\mu \le \int \limsup_n f_n\,d\mu\).
Dominated Convergence Theorem follows from Fatou plus reverse Fatou applied simultaneously to a sequence converging a.e. and dominated by an integrable \(h\): the liminf and limsup of \(\int f_n\) sandwich to a common limit equal to \(\int f\).
Fatou for measures / weak-* lower semicontinuity: if \(\mu_n \to \mu\) in an appropriate sense and \(f\ge0\) is lower semicontinuous, an analogous liminf inequality holds for varying measures, not just varying integrands — this is the measure-theoretic seed of lower semicontinuity results in the calculus of variations.
The converse is false: strict inequality is common (see Worked example 1), so equality cannot be deduced from the hypotheses alone; equality requires extra structure such as uniform integrability or a.e. convergence plus domination (Dominated/Vitali Convergence).

Fails without

Non-negativity dropped: on \( (\mathbb{R}, \mathcal{L}, \text{Leb}) \) take \( f_n = -\mathbf{1}_{[n, n+1]} \). Then \( f_n \to 0 \) pointwise everywhere, so \( \int \liminf_n f_n \, d\mu = \int 0\,d\mu = 0 \). But \( \int_{\mathbb{R}} f_n\,d\mu = -1 \) for every \(n\), so \( \liminf_n \int f_n\,d\mu = -1 \). Now \(0 \le -1\) is false: the inequality is violated.
Measurability dropped: let \(V \subseteq [0,1]\) be a Vitali set (non-measurable, exists via the Axiom of Choice) and set \(f_n = \mathbf{1}_V\) for every \(n\). Then \(\liminf_n f_n = \mathbf{1}_V\) is non-measurable, so \(\int \liminf_n f_n\,d\mu\) is simply undefined — the statement of the theorem cannot even be written down, let alone proved.
Countable additivity dropped (finitely additive content instead of a measure): Monotone Convergence — the load-bearing lemma in Step 4 — can fail for merely finitely additive set functions, and with it Fatou's inequality can fail; countable additivity of \(\mu\) is used essentially, not for convenience.

Common errors

Writing "\(=\)" instead of "\(\le\)" — students routinely misremember Fatou as an equality or silently assume the sequence converges, which is never given and not needed.
Applying Fatou directly to a sequence that takes negative values without first shifting by a dominating/minorizing function to restore non-negativity.
Confusing the direction with reverse Fatou: writing \(\int \limsup f_n \le \limsup \int f_n\) (wrong direction — the true reverse-Fatou inequality is \(\ge\), and it additionally requires a dominating integrable majorant, unlike the base lemma).
Forgetting that \(\liminf_n f_n\) must first be shown measurable before its integral is meaningful; students assume measurability "because each \(f_n\) is measurable" without citing the countable-operations closure lemma.
Trying to apply Fatou termwise to a series \(\sum_n a_n\) by treating partial sums naively without checking non-negativity of the summands \(a_n\), then being surprised the inequality fails for signed series.
Believing Fatou requires \(\mu(X) \lt \infty\) or \(\sigma\)-finiteness — it requires neither.

Discussion

Fatou's lemma was first proved by Pierre Fatou in 1906, in the course of his work on Fourier series and complex analysis (the same Fatou of Fatou sets in complex dynamics), predating Lebesgue's own systematic development of dominated convergence arguments and giving the field one of its most quietly load-bearing inequalities. Its logical position is foundational: rather than being a corollary of Monotone Convergence, in most modern treatments (as here) it is proved directly from MCT and then used to derive both directions of Dominated Convergence, making it the pivot between the "easy" monotone case and the general case.

The deep reason Fatou only gives an inequality, not an equality, is a phenomenon called "loss of mass in the limit" — probability mass, \(L^1\) norm, or measure can escape to infinity, spread out and thin to zero pointwise while ceasing to be summable, or concentrate on a null set. Each worked example below realizes a different escape mechanism (translation/traveling bump vs. concentration/spike), and this taxonomy is the standard way analysts diagnose why Dominated Convergence needs its extra domination hypothesis: domination is precisely the condition that rules out mass loss, converting Fatou's inequality into an equality.

In probability theory Fatou's lemma is usually stated for a sequence of non-negative random variables \(X_n\) on a probability space \((\Omega, \mathcal{F}, \mathbb{P})\): \( \mathbb{E}[\liminf_n X_n] \le \liminf_n \mathbb{E}[X_n] \). This underlies, for instance, the fact that the expected value of a non-negative supermartingale limit is at most the initial expectation, and is a standard tool for proving integrability of almost-sure limits of non-negative random variables without needing to control the whole sequence.

A conditional (measure-varying) version of Fatou plays a central role in optimal transport and the calculus of variations, where one shows that an energy or cost functional is lower semicontinuous with respect to weak convergence of measures: \( \int f \, d\mu \le \liminf_n \int f_n \, d\mu_n \) under joint convergence \(f_n \to f\) and \(\mu_n \rightharpoonup \mu\), a genuine generalization requiring more delicate hypotheses (e.g. lower semicontinuity of \(f\), tightness of \((\mu_n)\)) than the classical single-measure statement proved above. Common misconception: that Fatou requires pointwise convergence of \((f_n)\) — it does not; \(\liminf_n f_n\) is defined and the inequality holds for every sequence of non-negative measurable functions, convergent or not.

Worked examples

Ex.1

\text{On } (\mathbb{R}, \mathcal{L}, \text{Leb}),\quad f_n := \mathbf{1}_{[n, n+1]}\ \text{ ("traveling bump")}

Non-negative, measurable (indicator of an interval); a standard example used to illustrate strict inequality in Fatou via mass escaping to infinity. A

\text{For each fixed } x \in \mathbb{R},\quad f_n(x) = 0 \text{ for all } n \gt x, \text{ so } \liminf_{n\to\infty} f_n(x) = 0

Any fixed \(x\) is eventually outside \([n, n+1]\) once \(n \gt x\), since the interval marches off to \(+\infty\); hence the pointwise limit (and so also the \(\liminf\)) is identically \(0\). A

\int_{\mathbb{R}} \liminf_n f_n \, d\mu = \int_{\mathbb{R}} 0 \, d\mu = 0 \qquad\text{while}\qquad \int_{\mathbb{R}} f_n \, d\mu = 1 \ \ \forall n \implies \liminf_n \int f_n \, d\mu = 1

Left side by Step 2; right side because \([n,n+1]\) has Lebesgue length \(1\) for every \(n\), so the sequence of integrals is constantly \(1\) and its liminf is \(1\). A

0 \;=\; \int_{\mathbb{R}} \liminf_n f_n \, d\mu \;\lt\; \liminf_n \int_{\mathbb{R}} f_n \, d\mu \;=\; 1

Reading. Fatou's inequality holds, strictly: unit mass escapes to infinity and is entirely lost in the pointwise limit, exactly the phenomenon Fatou only bounds from one side.

Ex.2

\text{On } (\mathbb{R}, \mathcal{L}, \text{Leb}),\quad f_n := n\,\mathbf{1}_{[0, 1/n]}\ \text{ ("concentrating spike")}

Non-negative, measurable; every \(f_n\) has the same total integral, but the spike concentrates onto a single point as \(n\to\infty\), illustrating a second, distinct mass-loss mechanism from Example 1. A

\text{For } x = 0:\ f_n(0) = n \to \infty. \quad \text{For } x \gt 0:\ f_n(x) = 0 \text{ once } n \gt 1/x. \quad \text{For } x \lt 0:\ f_n(x)=0\ \forall n.

Direct evaluation of \(f_n(x) = n\,\mathbf 1_{[0,1/n]}(x)\) in each of the three cases; the interval \([0,1/n]\) shrinks to \(\{0\}\) as \(n \to \infty\). A

\liminf_{n\to\infty} f_n(x) = 0 \ \text{ for every } x \in \mathbb{R} \quad\Longrightarrow\quad \int_{\mathbb{R}} \liminf_n f_n\,d\mu = 0

From Step 2, \(f_n(x) \to 0\) for every \(x \ne 0\), and even at \(x=0\) the value \(f_n(0)=n \to \infty\) does not stop the \(\liminf\) definition from applying pointwise at each \(x\) separately; the single point \(x=0\) has Lebesgue measure zero and in any case the pointwise limit there is \(+\infty\), not part of a "liminf over \(n\) at fixed \(x\)" contradiction — the function \(\liminf_n f_n\) is \(0\) a.e., hence its integral is \(0\) by monotonicity/null-set invariance of the integral. B

\int_{\mathbb{R}} f_n \, d\mu = n \cdot \tfrac{1}{n} = 1 \quad \text{for every } n \implies \liminf_n \int_{\mathbb{R}} f_n\,d\mu = 1

Direct computation of the integral of a scaled indicator: height \(n\) times width \(1/n\). The resulting constant sequence \(1,1,1,\ldots\) has liminf \(1\). A

0 \;=\; \int_{\mathbb{R}} \liminf_n f_n \, d\mu \;\lt\; \liminf_n \int_{\mathbb{R}} f_n \, d\mu \;=\; 1

Reading. Again strict inequality, but here the unit mass is lost not by escaping to infinity but by concentrating onto the single Lebesgue-null point \(\{0\}\), which the integral of the (a.e. zero) limit function cannot see.

Problems

Let \((X,\mathcal M,\mu)\) be a measure space and \(f_n:X\to[0,\infty]\) measurable with \(f_n \to f\) pointwise \(\mu\)-a.e. Show \(\int_X f\,d\mu \le \liminf_n \int_X f_n\,d\mu\).
Solution
Since \(f_n \to f\) a.e., we have \(f = \liminf_n f_n\) a.e. (the liminf of a pointwise-convergent sequence equals the limit wherever convergence holds). Integrals are unaffected by modification on a null set, so \(\int_X f\,d\mu = \int_X \liminf_n f_n \, d\mu\). Apply Fatou's lemma directly to \((f_n)\) to get \(\int_X \liminf_n f_n\,d\mu \le \liminf_n \int_X f_n\,d\mu\), and chain the two equalities/inequality together.
Let \(g_n = \tfrac1n \mathbf{1}_{[0,n]}\) on \((\mathbb{R},\mathcal L,\text{Leb})\). Compute \(\int \liminf_n g_n\,d\mu\) and \(\liminf_n \int g_n\,d\mu\), and verify Fatou's inequality explicitly.
Solution
For each fixed \(x\), once \(n \gt \max(x,0)\) we have \(x \in [0,n]\), so \(g_n(x) = 1/n \to 0\); hence \(\liminf_n g_n = 0\) pointwise everywhere and \(\int_{\mathbb R} \liminf_n g_n\,d\mu = 0\). Meanwhile \(\int_{\mathbb R} g_n\,d\mu = \tfrac1n \cdot n = 1\) for every \(n\), so \(\liminf_n \int g_n\,d\mu = 1\). Indeed \(0 \le 1\), consistent with (and strict in) Fatou's lemma; mass escapes to infinity while simultaneously flattening to height \(0\), a hybrid of the two mechanisms in the worked examples.
Use Fatou's lemma to prove: if \(f_n \ge 0\) measurable and \(f_n \to f\) pointwise a.e. with \(\int_X f_n\,d\mu \le M\) for all \(n\) (some fixed constant \(M\)), then \(f\) is integrable with \(\int_X f\,d\mu \le M\).
Solution
By Problem 1 (equivalently, directly by Fatou applied to \(f_n\)), \(\int_X f\,d\mu \le \liminf_n \int_X f_n\,d\mu\). Since \(\int_X f_n\,d\mu \le M\) for every \(n\), the sequence \((\int_X f_n\,d\mu)_n\) is bounded above by \(M\), so its liminf is also \(\le M\). Hence \(\int_X f\,d\mu \le M \lt \infty\), so \(f\) is integrable (in \(L^1\), as \(f \ge 0\) a.e.).
Give an example of a sequence \(f_n \ge 0\) of measurable functions on \(([0,1],\mathcal L, \text{Leb})\) with \(f_n \to 0\) pointwise everywhere but \(\int_0^1 f_n \, d\mu \to \infty\), and reconcile this with Fatou's lemma.
Solution
Take \(f_n = n^2 \mathbf 1_{(0,1/n)}\). For fixed \(x \in (0,1]\), once \(n \gt 1/x\) we get \(f_n(x)=0\), so \(f_n(x) \to 0\); at \(x=0\), \(f_n(0)=0\) for all \(n\) too (open interval). So \(f_n \to 0\) pointwise on all of \([0,1]\), giving \(\liminf_n f_n = 0\) and \(\int_0^1 \liminf_n f_n\,d\mu = 0\). But \(\int_0^1 f_n\,d\mu = n^2 \cdot \tfrac1n = n \to \infty\), so \(\liminf_n \int f_n\,d\mu = \infty\). Fatou only claims \(0 \le \infty\), which is true and no contradiction — Fatou never claims the two sides are comparable in the other direction or bounded.
(Fatou for series) Let \(a_{n,k} \ge 0\) for all \(n,k \in \mathbb{N}\), viewed as functions of \(k\) on \((\mathbb N, \mathcal P(\mathbb N), \text{counting measure})\), i.e. \(f_n(k) := a_{n,k}\). Deduce from Fatou's lemma that \(\sum_{k=1}^\infty \liminf_{n\to\infty} a_{n,k} \ \le\ \liminf_{n\to\infty}\sum_{k=1}^\infty a_{n,k}\).
Solution
On \((\mathbb N, \mathcal P(\mathbb N), \#)\), every subset is measurable and integration against counting measure is exactly summation: \(\int_{\mathbb N} f_n \, d\# = \sum_{k=1}^\infty f_n(k) = \sum_{k=1}^\infty a_{n,k}\), and likewise \(\int_{\mathbb N}\liminf_n f_n\,d\# = \sum_{k=1}^\infty \liminf_n a_{n,k}\) (using \((\liminf_n f_n)(k) = \liminf_n a_{n,k}\) pointwise in \(k\)). Since each \(f_n = a_{n,\cdot} \ge 0\) is trivially measurable (every function on \((\mathbb N,\mathcal P(\mathbb N))\) is), Fatou's lemma applies verbatim to give \(\int_{\mathbb N} \liminf_n f_n \, d\# \le \liminf_n \int_{\mathbb N} f_n\,d\#\), which unwinds exactly to the claimed inequality for double-indexed non-negative sequences.