Fatou's Lemma for Indicator Sequences

From Fatou’s Lemma to a Set Inequality

For a sequence of measurable sets, the liminf records points that belong to every set from some point onward. Fatou’s lemma turns this idea into a useful measure inequality: the measure of the points that eventually remain in the sets cannot exceed the lower limit of the measures of the individual sets. The inequality does not require the sets to be nested, and it remains valid when some or all of the measures are infinite.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; no finiteness, sigma-finiteness, or completeness assumption is imposed. Each $E_n$ belongs to $\mathcal{M}$. The indicator $\chi_{E_n}:X\to\{0,1\}$ is a real-valued measurable function, and $\int_X\chi_{E_n}\,d\mu=\mu(E_n)$, with $+\infty$ allowed. We use Fatou’s lemma in its established form: for nonnegative measurable functions $f_n:X\to[0,\infty]$, $\int_X\liminf_n f_n\,d\mu\le\liminf_n\int_X f_n\,d\mu$.

Definition: The liminf of a sequence $(E_n)$ of sets is $$ \liminf_{n\to\infty}E_n=\bigcup_{m=1}^{\infty}\bigcap_{n=m}^{\infty}E_n. $$ Thus $x\in\liminf_n E_n$ exactly when there is an index $m$ such that $x\in E_n$ for every $n\ge m$. In words, $x$ belongs to all but finitely many of the sets.

Because $\mathcal{M}$ is a sigma-algebra, it is closed under countable unions and countable intersections. Therefore $\liminf_n E_n\in\mathcal{M}$, so its measure is defined. The key link to Fatou’s lemma is that the indicators have a pointwise liminf equal to the indicator of this set.

\liminf_{n\to\infty}\chi_{E_n}(x) = \begin{cases} 1,&x\in\liminf_{n\to\infty}E_n,\\ 0,&x\notin\liminf_{n\to\infty}E_n. \end{cases}

Indeed, a sequence of zeros and ones has liminf $1$ precisely when its terms are eventually all $1$. If it is not eventually all $1$, then it has infinitely many zero terms, so its liminf is $0$. Applying Fatou’s lemma to $f_n=\chi_{E_n}$ now gives the measure inequality.

The Measure Inequality

Theorem (Fatou’s lemma for indicator sequences): On any measure space $(X,\mathcal{M},\mu)$, for every sequence $(E_n)$ in $\mathcal{M}$, $$ \mu\!\left(\liminf_{n\to\infty}E_n\right) \le \liminf_{n\to\infty}\mu(E_n). $$ Both sides are interpreted in $[0,\infty]$. No assumption that the measures are finite is needed.

Proof. Set $f_n=\chi_{E_n}$. These functions are nonnegative and measurable, so Fatou’s lemma applies. The pointwise identity above gives $\liminf_n f_n=\chi_{\liminf_n E_n}$ everywhere. Using the indicator integral formula on both sides of Fatou’s inequality yields $$ \mu\!\left(\liminf_n E_n\right) =\int_X\chi_{\liminf_n E_n}\,d\mu =\int_X\liminf_n f_n\,d\mu \le\liminf_n\int_X f_n\,d\mu =\liminf_n\mu(E_n). $$ All terms may be infinite; the inequality is valid for extended nonnegative values. $\square$

The same inequality has a direct set-theoretic proof, which makes the role of eventual membership especially visible. For each $m\ge1$, define the measurable tail intersection $F_m=\bigcap_{n=m}^{\infty}E_n$. Then $F_m\subseteq F_{m+1}$ and $\bigcup_mF_m=\liminf_n E_n$. Continuity from below for measures, which applies without a finiteness hypothesis, says $\mu(\bigcup_mF_m)=\lim_m\mu(F_m)$. For every $n\ge m$, $F_m\subseteq E_n$, hence $\mu(F_m)\le\mu(E_n)$. Taking the infimum over $n\ge m$ gives $\mu(F_m)\le\inf_{n\ge m}\mu(E_n)$. The right-hand tail infima increase to $\liminf_n\mu(E_n)$, proving the stated bound. This argument also covers empty sets and infinite measures.

Worked Sequences

Worked Example: Alternating Halves of the Unit Interval

Let $X=[0,1]$ with the restricted Lebesgue sigma-algebra and measure, and define $E_n=[0,\tfrac12]$ for even $n$, while $E_n=[\tfrac12,1]$ for odd $n$. The sets are measurable. Each has measure $\tfrac12$, since each is a closed interval of length $\tfrac12$.

Every tail $\{E_n:n\ge m\}$ contains sets of both types. Their intersection is exactly $\{\tfrac12\}$: the common endpoint belongs to both closed intervals, while every other point is absent from one of them. Thus $F_m=\{\tfrac12\}$ for every $m$, and $$ \liminf_n E_n=\bigcup_{m=1}^{\infty}F_m=\{\tfrac12\}. $$ A singleton has Lebesgue measure zero, so the left side of the inequality is $0$. The sequence of measures is constantly $\tfrac12$, so its liminf is $\tfrac12$. Therefore $0\le\tfrac12$. Check: the only point that remains in every set in any tail is the common endpoint, and that endpoint has measure zero.

Worked Example: Intervals That Keep Moving

Take $X=\mathbb{R}$ with its Lebesgue sigma-algebra and Lebesgue measure $m$, and set $E_n=(n,n+1)$ for $n\ge1$. Each interval is measurable and has measure $1$. The intervals are pairwise disjoint, so any fixed $x\in\mathbb{R}$ belongs to at most one $E_n$. In particular, no $x$ belongs to every set in a tail. Hence $\liminf_n E_n=\varnothing$, whose measure is $0$, while $\liminf_n m(E_n)=\lim_n1=1$.

The inequality reads $0\le1$. Check: the empty set has measure zero, and substitution of $m(E_n)=1$ for every $n$ gives lower limit $1$. The fact that each individual set has positive measure does not force any point to remain in the sets eventually.

Worked Example: A Persistent Core and Moving Pieces

On $\mathbb{R}$ with Lebesgue measure $m$, let $$ E_n=[0,1]\cup[2n,2n+1],\qquad n\ge1. $$ The two intervals in each union are disjoint, each has length $1$, and so $m(E_n)=1+1=2$ for every $n$. Every $x\in[0,1]$ belongs to every $E_n$. A point outside $[0,1]$ can belong to at most one moving interval $[2n,2n+1]$, since these intervals are pairwise disjoint. It therefore cannot belong to every set in any tail. This proves $\liminf_n E_n=[0,1]$.

Consequently, $m(\liminf_nE_n)=m([0,1])=1$, whereas $\liminf_n m(E_n)=2$. The inequality is strict: $1<2$. Check: the persistent interval contributes length $1$; for each fixed $n$, the separate moving interval contributes another length $1$, giving $m(E_n)=2$. The moving contribution does not persist at any fixed point.

Worked Example: An Initial Exception Does Not Change the Liminf

On $X=[0,1]$ with restricted Lebesgue measure, let $E_1=\varnothing$ and $E_n=[0,1]$ for every $n\ge2$. For $m=1$, the tail intersection $\bigcap_{n\ge1}E_n$ is empty because it includes $E_1$. For every $m\ge2$, the tail intersection is $[0,1]$. Taking the union over all $m$ gives $\liminf_nE_n=[0,1]$, of measure $1$. The measures are $0,1,1,\ldots$, whose liminf is $1$. Thus both sides equal $1$.

Check: changing one initial set can change an individual tail intersection, but the union over all tail starting indices includes the tail beginning at $m=2$. The eventual membership condition is satisfied by every point of $[0,1]$.

What the Inequality Does and Does Not Say

The theorem gives a lower bound, not an equality in general. In the persistent-core example, each set has measure $2$, but only a set of measure $1$ persists. The extra measure can be carried by pieces that move from one set to another. This is why one cannot replace the inequality by an equality without additional information about the sequence.

It is also important to use the correct set operation. The liminf is a union of tail intersections, not the intersection of all the sets. For the sequence with $E_1=\varnothing$ and $E_n=[0,1]$ for $n\ge2$, the intersection $\bigcap_{n=1}^{\infty}E_n$ is empty, but the liminf is $[0,1]$. Confusing these sets would incorrectly discard points that belong to every sufficiently late set.

Takeaway: To apply Fatou’s lemma to sets, use $f_n=\chi_{E_n}$. The pointwise liminf of these indicators is the indicator of eventual membership, and the indicator integral formula converts Fatou’s inequality exactly into $$ \mu(\liminf_nE_n)\le\liminf_n\mu(E_n). $$

Check Your Understanding

For each question, identify the tail behavior of points before comparing measures.

On $\mathbb{R}$ with Lebesgue measure, let $E_n=(0,1/n)$. Determine $\liminf_nE_n$ and compare its measure with $\liminf_n m(E_n)$.
Let $E_n=[0,1]$ for even $n$ and $E_n=\varnothing$ for odd $n$, on $[0,1]$ with Lebesgue measure. Find the liminf set and the lower limit of the measures.
Explain why the Fatou inequality for $\chi_{E_n}$ remains valid if $\mu(E_n)=+\infty$ for every $n$.
Give a sequence for which $\bigcap_{n=1}^{\infty}E_n$ differs from $\liminf_nE_n$, and compute both sets.
In the persistent-core example, identify which part of each $E_n$ contributes to the liminf and which part does not.