Nonnegative Series Under the Integral

From Partial Sums to an Integral

A series of functions is defined pointwise through its partial sums. For nonnegative terms, those partial sums increase, which is exactly the structure needed to apply the Monotone Convergence Theorem. This gives a useful identity even when the measure space has infinite measure, the individual integrals are infinite, or the final answer is $+\infty$.

Throughout, let $(X,\mathcal{M},\mu)$ be any measure space; no finiteness or $\sigma$-finiteness assumption is imposed. Let $g_k:X\to[0,\infty]$ be measurable for each positive integer $k$. Define the partial sums $S_N(x)=\sum_{k=1}^{N}g_k(x)$, and define the pointwise series by $S(x)=\sum_{k=1}^{\infty}g_k(x)=\lim_{N\to\infty}S_N(x)$, with $+\infty$ allowed. Each $S_N$ is measurable, and $S_N\le S_{N+1}$ everywhere.

Theorem (Nonnegative series under the integral): For the measure space and measurable functions just specified, $$ \int_X\left(\sum_{k=1}^{\infty}g_k\right)\,d\mu = \sum_{k=1}^{\infty}\int_X g_k\,d\mu. $$ Both sides are defined in $[0,\infty]$, and equality includes the possibility that both sides are infinite.

Proof. The sequence $(S_N)$ consists of measurable, nonnegative functions and increases pointwise to the measurable function $S$. The Monotone Convergence Theorem, applied on $(X,\mathcal{M},\mu)$, therefore gives $\int_X S\,d\mu=\lim_{N\to\infty}\int_X S_N\,d\mu$. For each finite $N$, additivity of the nonnegative integral gives $\int_X S_N\,d\mu=\sum_{k=1}^{N}\int_X g_k\,d\mu$. Taking the limit of these increasing partial sums yields the stated identity, with $+\infty$ permitted. $\square$

The theorem does not assert that the common value is finite. It also does not require the measure space to be finite, or each $g_k$ to have a finite integral. The essential hypothesis is that every term is nonnegative, so that the partial sums increase. The Monotone Convergence Theorem is the standard result used here; its proof is part of the earlier material on that theorem.

Examples with a Finite Common Value

Worked Example: A Factorial Series on the Unit Interval

Take Lebesgue measure $m$ on $[0,1]$, and let $g_k(x)=x/k!$ for $k\ge1$. These are measurable and nonnegative on the whole interval, including its endpoints. Compute the integral of their pointwise sum.

For each $N$, the partial sum is $S_N(x)=x\sum_{k=1}^{N}1/k!$. The functions $S_N$ increase to $S(x)=x\sum_{k=1}^{\infty}1/k!=x(e-1)$, using the standard power-series expansion $e=\sum_{k=0}^{\infty}1/k!$. Each term has integral

\int_0^1 g_k(x)\,dx =\frac{1}{k!}\int_0^1x\,dx =\frac{1}{2k!}.

Consequently, the theorem gives $$ \int_0^1 S(x)\,dx =\sum_{k=1}^{\infty}\frac{1}{2k!} =\frac{e-1}{2}. $$ Here the hypotheses have been checked: all $g_k$ are measurable and nonnegative, and their partial sums increase pointwise. Check: integrating the pointwise sum directly gives $\int_0^1 x(e-1)\,dx=(e-1)/2$, since $\int_0^1x\,dx=1/2$. The first three term integrals are $1/2,\ 1/4,\ 1/12$, obtained respectively from $1/(2\cdot1!),1/(2\cdot2!),1/(2\cdot3!)$.

In this example, both sides are finite. More generally, the identity itself decides whether the integral is finite: $\int_X S\,d\mu<\infty$ if and only if the numerical series $\sum_k\int_Xg_k\,d\mu$ converges to a finite value. This equivalence follows directly from equality of the two extended nonnegative quantities; it does not require a separate integrability assumption on $S$.

Examples with an Infinite Common Value

Worked Example: A Finite Pointwise Sum with Infinite Integral

On $[0,1]$ with Lebesgue measure, set $g_n=\chi_{(0,1/n)}$ for each $n\ge1$. The interval is open: a point $x$ belongs to the $n$-th set exactly when $0<x<1/n$. Each $g_n$ is measurable and nonnegative, and $\int_0^1g_n\,dm=1/n$. Therefore the series of integrals is the harmonic series, which diverges to $+\infty$.

We can also see directly why the pointwise sum has infinite integral. At $x=0$, every term is zero. At any $x>0$, only the positive integers $n<1/x$ contribute, so $S(x)$ is finite at each such point. For $0<x<1/2$, however, the number of positive integers strictly less than $1/x$ is at least $1/x-1$, and $1/x-1\ge1/(2x)$ because $x<1/2$. Thus $S(x)\ge1/(2x)$ on $(0,1/2)$. For every $a\in(0,1/2)$,

\int_0^{1/2}S(x)\,dx \ge \int_a^{1/2}\frac{1}{2x}\,dx =\frac12\log\left(\frac{1}{2a}\right).

The right-hand side is unbounded as $a$ decreases to zero, so $\int_0^1S\,dm=+\infty$. This agrees with the series of integrals. In this example, the pointwise sum is finite at every point, but its integral is infinite.

Worked Example: Infinite Values on an Infinite-Measure Space

Take Lebesgue measure $m$ on $\mathbb{R}$, and for each $n\ge1$ define $g_n=\chi_{[n,n+1)}$. These half-open intervals are pairwise disjoint, and each has measure $1$. Hence $\int_{\mathbb{R}}g_n\,dm=1$, so $\sum_{n=1}^{\infty}\int_{\mathbb{R}}g_n\,dm=+\infty$.

For a point $x\in[n,n+1)$, exactly one term equals $1$ and all other terms equal $0$; for $x<1$, all terms equal $0$. The pointwise sum is therefore $\chi_{[1,\infty)}$. Its integral is $m([1,\infty))=+\infty$. As a check, each partial sum is the indicator of $[1,N+1)$, whose integral is the interval length $N$. These partial-sum integrals increase without bound, in agreement with the theorem.

Why Nonnegativity Is Essential

The result cannot be applied to a series with signed terms merely because each term is integrable. For a concrete example, work on $\mathbb{R}$ with Lebesgue measure and define $f_n=\chi_{[n,n+1)}-\chi_{[n+1,n+2)}$. Each $f_n$ is measurable and integrable, and $\int_{\mathbb{R}} f_n\,dm=1-1=0$. Thus the series of term integrals is $0$.

But the pointwise series has a different sum. On $[1,2)$, only the first positive indicator contributes, so the sum is $1$. On every interval $[j,j+1)$ for integer $j\ge2$, the positive indicator from $f_j$ and the negative indicator from $f_{j-1}$ cancel, so the sum is $0$. Outside $[1,\infty)$, every term is zero. Hence $\sum_{n=1}^{\infty}f_n=\chi_{[1,2)}$ pointwise, and its integral is $1$, not $0$. The partial sums are not an increasing sequence of nonnegative functions, so the Monotone Convergence Theorem does not apply.

Common mistake: Writing $\int\sum_n f_n\,d\mu=\sum_n\int f_n\,d\mu$ for arbitrary signed terms and justifying it by monotone convergence. The calculation above would give $1=0$, a contradiction. The correction is to check that the summands are nonnegative and measurable before using this theorem. Signed series require other hypotheses and results; nonnegative monotone convergence supplies no interchange rule for them.

How to Apply the Identity

For a proposed series under an integral, make the relevant checks in this order:

Specify the measure space and verify that each term $g_k$ is measurable on it.
Verify pointwise nonnegativity, including at endpoints or exceptional points, so that the partial sums increase.
Define the pointwise sum as the limit of those partial sums, allowing the value $+\infty$.
Apply monotone convergence to the partial sums, then evaluate the finite partial-sum integrals and take their increasing limit.

No separate assumption that the terms have finite integrals is needed: if some term has infinite integral, or the sum of the term integrals diverges, the theorem states that the integral of the pointwise sum is infinite as well. Conversely, when the series of integrals has a finite sum, the integral of the pointwise sum is finite and equals that sum.

Check Your Understanding

For each question, identify the measure space and check the nonnegativity and measurability assumptions before using the series identity.

On $[0,1]$ with Lebesgue measure, take $g_k(x)=x/k!$. What is $\int_0^1\sum_{k=1}^{\infty}g_k(x)\,dx$, and which theorem justifies passing from partial sums to the limit?
On $[0,1]$, take $g_n=\chi_{(0,1/n)}$. Is the pointwise sum finite at every point? Is its integral finite? Explain the distinction.
On $\mathbb{R}$, take $g_n=\chi_{[n,n+1)}$. Compute the integral of the $N$-th partial sum and use it to determine the limiting integral.
For $f_n=\chi_{[n,n+1)}-\chi_{[n+1,n+2)}$, compute the pointwise series sum and the series of integrals. Which hypothesis needed for monotone convergence is missing?