Fatou's Lemma

Lower Limits and Integrals

The Monotone Convergence Theorem applies when measurable functions increase pointwise. In many sequences, however, the functions themselves do not increase. The pointwise lower limit still records values that persist arbitrarily far along the sequence, and Fatou’s Lemma relates that lower limit to the integrals of the sequence.

The lemma is an inequality, not an assertion that integration always commutes with a pointwise limit. In particular, it gives a lower bound for the lower limit of the integrals; it does not by itself give an upper bound or an equality. The distinction matters when mass moves to new parts of the space or becomes concentrated on smaller sets.

The Pointwise Liminf

For a sequence of extended nonnegative real numbers $a_n$, its lower limit is the limit of the infima of its tails. Those tail infima increase as the tails become shorter, so the limit exists in $[0,\infty]$. For measurable functions, the same construction is performed pointwise.

Definition: If $f_n:X\to[0,\infty]$ is a sequence of functions, define $$ \liminf_{n\to\infty} f_n(x) = \lim_{n\to\infty}\inf_{k\geq n} f_k(x). $$ The value is the smallest extended value that the sequence approaches arbitrarily far along its tail. If every $f_n$ is measurable, this pointwise liminf is measurable, by the Measurability of Limsup and Liminf theorem.

The liminf need not be the ordinary pointwise limit: the sequence may fail to converge. For example, a sequence that alternates between $0$ and $3$ has liminf $0$ and limsup $3$. If the sequence does converge, its liminf equals its limit.

Fatou’s Lemma

Theorem (Fatou’s Lemma): Let $(X,\mathcal{F},\mu)$ be a measure space, and let $f_n:X\to[0,\infty]$ be measurable for every positive integer $n$. Then $$ \int_X \liminf_{n\to\infty} f_n\,d\mu \leq \liminf_{n\to\infty}\int_X f_n\,d\mu. $$ Both sides are interpreted in $[0,\infty]$. No finiteness assumption on the measure space or on the integrals is required.

The proof of Fatou’s Lemma is the subject of the next tutorial. Its key structure is that the tail infima of the functions form an increasing sequence, so the Monotone Convergence Theorem can be used. Here we focus on reading the conclusion correctly and applying it.

If the right side is finite, the lemma says that the integral of the pointwise liminf cannot exceed the limiting lower value of the integrals. If the right side is infinite, the inequality remains valid but may provide no finite bound. The hypotheses allow $f_n$ to take the value $+\infty$, and either side of the inequality may be infinite.

Worked Example: Functions Concentrating Near the Origin

On $[0,1]$ with Lebesgue measure, define $$ f_n(x)=n\mathbf{1}_{(0,1/n)}(x). $$ Each $f_n$ is nonnegative and measurable. At $x=0$, every $f_n(0)=0$. If $x>0$, choose an integer $N>1/x$. For $n\geq N$, we have $1/n<x$, so $x\notin(0,1/n)$ and $f_n(x)=0$. Thus $f_n(x)\to0$ at every point of $[0,1]$, and $$ \int_{[0,1]}\liminf_{n\to\infty}f_n\,d\lambda=0. $$ On the other hand, the interval $(0,1/n)$ has measure $1/n$, so $$ \int_{[0,1]}f_n\,d\lambda =n\lambda((0,1/n)) =n\cdot\frac{1}{n} =1. $$ Consequently, Fatou’s inequality reads $0\leq1$. The integrals do not tend to the integral of the pointwise limit: the functions become taller while their supports shrink.

Worked Example: An Oscillating Sequence

Let $X=[0,1]$ with Lebesgue measure, and set $f_n(x)=0$ when $n$ is even and $f_n(x)=2$ when $n$ is odd. For every $x$, the sequence alternates between $0$ and $2$, so $$ \liminf_{n\to\infty}f_n(x)=0. $$ Its integral is therefore zero. The integrals of the functions alternate between $0$ and $2$, whose lower limit is $0$. In this case Fatou’s inequality is an equality: $$ \int_{[0,1]}\liminf_{n\to\infty}f_n\,d\lambda =0 =\liminf_{n\to\infty}\int_{[0,1]}f_n\,d\lambda. $$ The pointwise sequence does not converge, but the liminf and its integral are still well-defined.

Consequences for Convergence and Sets

When $f_n$ converges pointwise everywhere to a measurable nonnegative function $f$, its liminf is $f$. Fatou’s Lemma then gives a particularly useful lower bound. The same conclusion holds for almost-everywhere convergence when the limit $f$ is measurable: the liminf equals $f$ outside a measurable null set, and changing a measurable nonnegative function on a null set does not change its integral.

Corollary (Lower Bound Under Pointwise Convergence): Suppose the nonnegative measurable functions $f_n$ converge pointwise everywhere to a measurable function $f:X\to[0,\infty]$. Then $$ \int_X f\,d\mu \leq \liminf_{n\to\infty}\int_X f_n\,d\mu. $$

Proof. Pointwise convergence gives $\liminf_n f_n(x)=f(x)$ for every $x$. Substitute this identity into Fatou’s Lemma to obtain the stated inequality.

There is also a set version. The liminf of a sequence of sets consists of points that belong to every set from some stage onward. Applying Fatou’s Lemma to indicator functions gives a lower bound on the measure of that eventual-membership set.

Theorem (Fatou’s Lemma for Sets): If $E_n\in\mathcal{F}$ for every positive integer $n$, then $$ \mu\left(\liminf_{n\to\infty}E_n\right) \leq \liminf_{n\to\infty}\mu(E_n), $$ where $\liminf_n E_n=\bigcup_{N=1}^{\infty}\bigcap_{n\geq N}E_n$.

Proof. Apply Fatou’s Lemma to the measurable functions $f_n=\mathbf{1}_{E_n}$. At a point $x$, the numerical sequence $\mathbf{1}_{E_n}(x)$ has liminf $1$ exactly when $x\in E_n$ for every sufficiently large $n$; otherwise its liminf is $0$. Therefore $$ \liminf_{n\to\infty}\mathbf{1}_{E_n} =\mathbf{1}_{\liminf_{n\to\infty}E_n}. $$ The Integral of an Indicator theorem identifies the integral of this indicator with the measure of the set. Fatou’s inequality now gives the claim.

Worked Example: Sets That Move Apart

On $\mathbb{R}$ with Lebesgue measure, let $E_n=[n,n+1]$. Each set has measure $1$. No real number belongs to $E_n$ for all sufficiently large $n$, because the intervals move arbitrarily far to the right. Hence $$ \liminf_{n\to\infty}E_n=\varnothing. $$ The set version of Fatou’s Lemma gives $$ 0=\lambda(\varnothing) \leq \liminf_{n\to\infty}\lambda(E_n) =1. $$ The inequality can be strict: the measures of the individual sets do not force their eventual-membership set to have positive measure.

Localizing the Inequality

Fatou’s Lemma can be applied on any measurable part of the space by multiplying each function by the indicator of that part. This is useful when the behavior of a sequence differs across regions or when only a local integral bound is needed.

Theorem (Localized Fatou Inequality): Let $A\in\mathcal{F}$, and suppose $f_n:X\to[0,\infty]$ are measurable. Then $$ \int_A\liminf_{n\to\infty}f_n\,d\mu \leq \liminf_{n\to\infty}\int_A f_n\,d\mu. $$

Proof. The functions $g_n=f_n\mathbf{1}_A$ are nonnegative and measurable. At every $x\in A$, $g_n(x)=f_n(x)$ for all $n$, so $\liminf_n g_n(x)=\liminf_n f_n(x)$. At every $x\notin A$, $g_n(x)=0$ for all $n$, and its liminf is zero. Thus $$ \liminf_{n\to\infty}g_n =\mathbf{1}_A\liminf_{n\to\infty}f_n. $$ Apply Fatou’s Lemma to $g_n$. By the definition of integration over $A$, the resulting inequality is exactly the localized statement. This argument also covers infinite integrals.

An Integrable Upper Bound: The Reverse Inequality

Fatou’s Lemma has the direction $\int\liminf f_n\leq\liminf\int f_n$. A useful reverse inequality is available when the sequence is bounded above by an integrable function and converges almost everywhere. Its hypotheses must be stated carefully: the limit must be measurable, and subtracting two extended values equal to $+\infty$ is not permitted.

Theorem (Reverse Fatou Inequality Under an Integrable Bound): Suppose $f_n,f,g:X\to[0,\infty]$ are measurable, $0\leq f_n\leq g$ everywhere for every $n$, $\int_X g\,d\mu<\infty$, and $f_n\to f$ almost everywhere. Then $$ \limsup_{n\to\infty}\int_X f_n\,d\mu \leq \int_X f\,d\mu. $$

Proof. Since $\int_Xg\,d\mu<\infty$, the set $Z=\{x:g(x)=+\infty\}$ is null. Indeed, for every positive integer $m$, $$ m\mu(Z)=\int_X m\mathbf{1}_Z\,d\mu \leq\int_X g\,d\mu<\infty, $$ so $\mu(Z)=0$. Let $N$ be a measurable null set outside which $f_n\to f$. On $X\setminus N$, the limit satisfies $f\leq g$. All the functions $f_n$ have finite integrals by monotonicity, since $\int f_n\leq\int g<\infty$. Also $f$ is integrable: outside $N$, it is bounded by $g$, and its integral on the null set $N$ is zero.

Define measurable nonnegative functions $$ h_n(x)= \begin{cases} g(x)-f_n(x),&x\notin Z,\\ 0,&x\in Z. \end{cases} $$ On $X\setminus Z$, $g$ is finite and $f_n\leq g$, so this subtraction is defined and nonnegative. On $Z$, the definition avoids the undefined expression $+\infty-(+\infty)$. Fatou’s Lemma applies to $h_n$.

Outside $N\cup Z$, the sequence $h_n$ converges to $g-f$. Define $q=g-f$ on $X\setminus(N\cup Z)$ and $q=0$ on $N\cup Z$. Then $q$ is measurable and nonnegative, and $\liminf_n h_n=q$ almost everywhere. Null-set changes do not affect these nonnegative integrals, so $$ \int_X\liminf_{n\to\infty}h_n\,d\mu =\int_Xq\,d\mu =\int_Xg\,d\mu-\int_Xf\,d\mu. $$ For each $n$, the pointwise identity $g=f_n+h_n$ holds off $Z$. Replacing $g$ and $f_n$ by zero on $Z$ leaves their integrals unchanged, since $Z$ is null; therefore finite additivity for nonnegative integrals gives $$ \int_Xh_n\,d\mu=\int_Xg\,d\mu-\int_Xf_n\,d\mu. $$ Fatou’s Lemma now implies $$ \int_Xg\,d\mu-\int_Xf\,d\mu \leq \liminf_{n\to\infty}\left(\int_Xg\,d\mu-\int_Xf_n\,d\mu\right) = \int_Xg\,d\mu-\limsup_{n\to\infty}\int_Xf_n\,d\mu. $$ All the integrals being subtracted are finite. Rearranging proves the result.

Worked Example: A Bounded Convergent Sequence

On $[0,1]$ with Lebesgue measure, let $f_n(x)=x^n$, $f(x)=0$ for $0\leq x<1$, and $f(1)=1$. Then $f_n\to f$ pointwise, and $0\leq f_n\leq1$, where the constant bound $g=1$ has integral $1$. Direct calculation gives $$ \int_{[0,1]}f_n\,d\lambda =\int_0^1x^n\,dx =\frac{1}{n+1} \longrightarrow 0. $$ The function $f$ is zero except at the single point $1$, so its integral is zero. Thus the reverse Fatou inequality reads $0\leq0$. The measurable-limit hypothesis holds here, including at the endpoint.

What Fatou’s Lemma Does Not Say

The lower bound from Fatou’s Lemma should not be mistaken for convergence of the integrals. In the concentration example, the functions converge pointwise to zero but every integral is $1$. The lemma correctly gives only $0\leq1$. Extra hypotheses are needed to conclude that the integrals converge to the integral of the limit.

A useful check before applying either inequality is to verify each of the following:

Each function to which Fatou’s Lemma is applied is measurable and nonnegative.
The pointwise liminf is the function whose integral appears on the left.
Any asserted limit function is measurable, especially when convergence is only almost everywhere.
In a reverse Fatou argument, the common upper bound has finite integral and any subtraction is defined on the region where it is used.

Key takeaway: Fatou’s Lemma gives a lower bound for the integral of a pointwise liminf. Under almost-everywhere convergence and an integrable upper bound, a separate reverse inequality gives an upper bound for the limsup of the integrals.

Check Your Understanding

Use the definitions and inequalities in this tutorial to answer these questions.

How is the pointwise liminf of $f_n(x)$ defined using infima over tails?
Which direction does Fatou’s Lemma give, and does it require the integrals to be finite?
For measurable sets $E_n$, what does membership in $\liminf_n E_n$ mean?
Why does the reverse Fatou inequality require the limit function to be measurable?
Why must the expression $g-f_n$ be defined separately on the set where $g=+\infty$?

What You'll Learn