Fatou's Lemma for Series and Counting Measure

From Counting Measure to Series

For a sequence of nonnegative terms, Fatou’s lemma compares the sum of the pointwise lower limits with the lower limit of the sums. The comparison is useful when the terms vary with an index and their sums are difficult to calculate exactly: it gives a lower bound using only the eventual behavior at each fixed index. It does not assert that the two quantities are equal.

Throughout, let $\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{P}(\mathbb{N})$ be its power set, and let $\#$ denote counting measure, so $\#(E)$ is the number of elements of $E$, possibly $+\infty$. The measure space $(\mathbb{N},\mathcal{P}(\mathbb{N}),\#)$ is complete and $\sigma$-finite, but not finite: $\mathbb{N}$ is a countable union of singletons of measure $1$, while $\#(\mathbb{N})=\infty$. Every function on this measurable space is measurable.

Definition: For a sequence $a_{n,k}\in[0,\infty]$, indexed by $n,k\in\mathbb{N}$, its $n$-th series is $\sum_{k=1}^{\infty}a_{n,k}$, defined as the limit of its increasing finite partial sums and allowed to equal $+\infty$. For each fixed $k$, $\liminf_{n\to\infty}a_{n,k}$ is the limit of the increasing sequence $\inf_{n\ge j}a_{n,k}$ as $j\to\infty$, also allowed to equal $+\infty$.

Define $f_n:\mathbb{N}\to[0,\infty]$ by $f_n(k)=a_{n,k}$. The integral of a nonnegative function with respect to counting measure is its series, including when the value is infinite:

\int_{\mathbb{N}} f_n\,d\#=\sum_{k=1}^{\infty}a_{n,k}.

Thus Fatou’s lemma on this measure space becomes an inequality about series. The result below states that inequality and proves it in the sequence notation itself. Its proof uses the Monotone Convergence Theorem, which applies to nonnegative measurable functions increasing pointwise; those hypotheses will be checked explicitly.

Fatou’s Inequality for Nonnegative Series

Theorem (Fatou’s lemma for series): Let $a_{n,k}\in[0,\infty]$ for all $n,k\in\mathbb{N}$. Then $$ \sum_{k=1}^{\infty}\liminf_{n\to\infty}a_{n,k} \ \le\ \liminf_{n\to\infty}\sum_{k=1}^{\infty}a_{n,k}. $$ Both sides are in $[0,\infty]$, and the inequality includes the cases where one or both sides are infinite.

Proof. For $j,k\in\mathbb{N}$, set $c_{j,k}=\inf_{n\ge j}a_{n,k}$. Since the set of indices $n\ge j+1$ is contained in the set of indices $n\ge j$, $c_{j,k}\le c_{j+1,k}$. For each fixed $k$, the limit of $c_{j,k}$ as $j\to\infty$ is $\liminf_{n\to\infty}a_{n,k}$. Each function $k\mapsto c_{j,k}$ is nonnegative and measurable on $(\mathbb{N},\mathcal{P}(\mathbb{N}),\#)$. The Monotone Convergence Theorem therefore gives $$ \sum_{k=1}^{\infty}\liminf_{n\to\infty}a_{n,k} = \lim_{j\to\infty}\sum_{k=1}^{\infty}c_{j,k}. $$ For every $n\ge j$ and every $k$, $c_{j,k}\le a_{n,k}$. Summing this pointwise inequality by monotonicity of the nonnegative integral gives $$ \sum_{k=1}^{\infty}c_{j,k} \le \sum_{k=1}^{\infty}a_{n,k} \qquad(n\ge j). $$ Consequently, $\sum_k c_{j,k}\le\inf_{n\ge j}\sum_k a_{n,k}$. Taking $j\to\infty$, the right side tends to $\liminf_{n\to\infty}\sum_k a_{n,k}$, by the definition of the lower limit of a sequence. Combining these inequalities proves the claim, including when a limit is infinite. $\square$

Corollary (finite-coordinate lower bound): Under the theorem’s hypotheses, for every finite set $F\subseteq\mathbb{N}$, $$ \sum_{k\in F}\liminf_{n\to\infty}a_{n,k} \le \liminf_{n\to\infty}\sum_{k=1}^{\infty}a_{n,k}. $$

Proof. The terms are nonnegative, so the sum over $F$ is at most the sum over all of $\mathbb{N}$. Apply the theorem to bound that full sum by the lower limit of the series totals. If $F$ is empty, its sum is $0$, and the inequality also holds because the right side is nonnegative. $\square$

This finite-coordinate form is often the practical first step: retain a finite collection of coordinates, take their pointwise lower limits, and then enlarge the collection. The infinite-series statement follows because the full nonnegative series is the supremum of its finite partial sums.

Computing the Lower Bound

Worked Example: A Single Unit Moves to Larger Indices

Set $a_{n,k}=1$ when $k=n$, and $a_{n,k}=0$ when $k\ne n$. Determine both sides of Fatou’s inequality.

Fix $k$. If $n>k$, then $n\ne k$, so $a_{n,k}=0$. Thus for every fixed $k$, the sequence $a_{n,k}$ is eventually zero and $\liminf_{n\to\infty}a_{n,k}=0$. Therefore $$ \sum_{k=1}^{\infty}\liminf_{n\to\infty}a_{n,k} =\sum_{k=1}^{\infty}0=0. $$ For each fixed $n$, exactly one term in the series over $k$ is $1$, namely the term with $k=n$, and all other terms are $0$. Hence $\sum_{k=1}^{\infty}a_{n,k}=1$ for every $n$, so its lower limit is $1$. Check: the resulting inequality is $0\le1$, as required. The pointwise lower limits record what happens at each fixed coordinate; they do not record that the unit term keeps appearing at a new coordinate.

Worked Example: A Geometric Tail Added to a Fixed Series

For $n,k\in\mathbb{N}$, let $$ a_{n,k}=2^{-k}\bigl(1+\chi_{\{k\ge n\}}(k)\bigr). $$ Here $\chi_{\{k\ge n\}}(k)$ is $1$ exactly when $k\ge n$, and is $0$ otherwise. Compute the pointwise lower limits and the lower limit of the series totals.

Fix $k$. For every $n>k$, the condition $k\ge n$ is false, so $a_{n,k}=2^{-k}$ for all sufficiently large $n$. Consequently, $\liminf_{n\to\infty}a_{n,k}=2^{-k}$, and $$ \sum_{k=1}^{\infty}\liminf_{n\to\infty}a_{n,k} =\sum_{k=1}^{\infty}2^{-k}=1. $$ For each $n$, the terms with $k<n$ sum to $\sum_{k=1}^{n-1}2^{-k}$, while the terms with $k\ge n$ sum to $2\sum_{k=n}^{\infty}2^{-k}$. Since $\sum_{k=n}^{\infty}2^{-k}=2^{1-n}$, the total is $$ \sum_{k=1}^{\infty}a_{n,k} =\sum_{k=1}^{n-1}2^{-k}+2\sum_{k=n}^{\infty}2^{-k} =(1-2^{1-n})+2^{2-n} =1+2^{1-n}. $$ It tends to $1$ as $n\to\infty$. Check: at $n=1$ the total is $1+2^0=2$; at $n=2$ it is $1+2^{-1}=3/2$; at $n=3$ it is $1+2^{-2}=5/4$. These agree with direct summation and approach the computed left side, $1$.

Worked Example: Truncated Geometric Series

Let $a_{n,k}=2^{-k}$ when $k\le n$, and $a_{n,k}=0$ when $k>n$. Find the pointwise lower-limit series and the lower limit of the totals.

For any fixed $k$, every $n\ge k$ satisfies $k\le n$, so $a_{n,k}=2^{-k}$ from that point onward. Therefore the pointwise lower limit is $2^{-k}$, and its series is $$ \sum_{k=1}^{\infty}\liminf_{n\to\infty}a_{n,k} =\sum_{k=1}^{\infty}2^{-k}=1. $$ For each $n$, the total is the finite geometric sum $$ \sum_{k=1}^{\infty}a_{n,k} =\sum_{k=1}^{n}2^{-k} =1-2^{-n}. $$ The identity follows by multiplying the finite sum by $2$ and subtracting the original sum. As $n\to\infty$, $2^{-n}\to0$, so the lower limit of the totals is $1$. Check: the totals at $n=1,2,3$ are $1/2$, $3/4$, and $7/8$, respectively, matching $1-2^{-n}$; both sides of Fatou’s inequality equal $1$.

How to Use the Inequality Safely

To apply the result, first identify the two indices: $n$ varies along the sequence, while $k$ indexes the terms within each series. Next, fix $k$ and calculate $\liminf_n a_{n,k}$. Only after doing this coordinate by coordinate should you sum those lower limits. Separately, calculate or bound the lower limit of the full series totals. The theorem compares these quantities in the stated direction; it does not permit exchanging the order of a limit and an infinite sum as an equality.

The nonnegativity assumption is essential to this formulation. It ensures that the series and integrals used in the proof are defined without cancellation, that finite partial sums increase to the full series, and that monotonicity applies to the sums. For signed terms, a series may have cancellation or fail to converge, so the argument above does not establish an analogous claim.

Common mistake: reversing the inequality. In the moving-unit example, every series total equals $1$, while every fixed coordinate has lower limit $0$. Reversing Fatou’s inequality would assert $1\le0$, which is false. The correct calculation is $$ \sum_{k=1}^{\infty}\liminf_{n\to\infty}a_{n,k}=0 \le 1=\liminf_{n\to\infty}\sum_{k=1}^{\infty}a_{n,k}. $$ The sum of coordinatewise lower limits is a lower bound for the lower limit of the totals, not an upper bound.

Fatou’s inequality is useful even when the series totals cannot be evaluated: any lower estimate for the pointwise lower-limit series is automatically a lower estimate for the lower limit of the totals. The finite-coordinate corollary provides such estimates using only finitely many terms, and the full inequality follows by considering all coordinates together.

Check Your Understanding

In each problem, keep the sequence index $n$ distinct from the series index $k$, and state which quantities may be infinite.

Suppose $a_{n,k}=3^{-k}$ for every $n,k$. Compute both sides of Fatou’s inequality.
For $a_{n,k}=1$ when $k=2n$ and $0$ otherwise, find the pointwise lower limits and the series total for each $n$.
Explain why $\sum_{k\in F}\liminf_n a_{n,k}$ is bounded above by the lower limit of the full series totals for every finite $F\subseteq\mathbb{N}$.
State precisely where nonnegativity is used in the proof of Fatou’s inequality for series.
Give an example where every series total is finite but the pointwise lower-limit series is infinite, or show that no such example can satisfy Fatou’s inequality.

Fatou's Lemma for Series and Counting Measure

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