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Tonelli for Countable Sums and Integrals

Learn when a nonnegative series can be integrated term by term, how counting measure turns integrals into sums, and what finiteness of the resulting integral implies.

Intermediate 10 min read

What You'll Learn

  • State the countable-sum form of Tonelli’s theorem for nonnegative measurable functions.
  • Prove the interchange using increasing partial sums and the monotone convergence theorem.
  • Apply the result to series of functions and compute an example with a finite integral.
  • Use counting measure to interchange the order of two nonnegative countable sums.
  • Deduce when a sum of nonnegative functions is finite almost everywhere and integrable.

From Finite Sums to Infinite Series

For finitely many nonnegative measurable functions, linearity gives the integral of their sum as the sum of their integrals. For an infinite sequence of such functions, the natural question is whether the same principle holds:

$$ \int_X \sum_{n=1}^{\infty} f_n\,d\mu \stackrel{?}{=} \sum_{n=1}^{\infty}\int_X f_n\,d\mu. $$

The answer is yes for nonnegative measurable functions. Both sides are allowed to equal \(+\infty\). This is an important feature: the equality does not require a prior assumption that the series or either integral is finite.

The result is the countable-sum form of Tonelli’s theorem. Its proof is a direct application of the monotone convergence theorem to the partial sums.

Theorem (Tonelli for countable sums): Let \((X,\mathcal{F},\mu)\) be a measure space, and let \(f_n:X\to[0,\infty]\) be measurable for each \(n\in\mathbb{N}\). Define the pointwise sum as the increasing limit of its partial sums. Then $$ \int_X \left(\sum_{n=1}^{\infty} f_n\right)\,d\mu = \sum_{n=1}^{\infty}\int_X f_n\,d\mu, $$ where either side may be infinite. No finiteness assumption on \(\mu\) is required.

Why the Interchange Is Valid

For each \(N\in\mathbb{N}\), define the partial sum \(S_N=\sum_{n=1}^N f_n\). Each \(S_N\) is measurable, and nonnegativity gives \(S_N\leq S_{N+1}\) at every point. By definition, the pointwise sum \(S=\sum_{n=1}^{\infty}f_n\) is \(\lim_{N\to\infty}S_N\).

Proof. The sequence \((S_N)\) is an increasing sequence of nonnegative measurable functions with pointwise limit \(S\). The monotone convergence theorem therefore gives $$ \int_X S\,d\mu = \lim_{N\to\infty}\int_X S_N\,d\mu. $$ For each finite \(N\), additivity of the integral for nonnegative functions gives $$ \int_X S_N\,d\mu = \sum_{n=1}^{N}\int_X f_n\,d\mu. $$ Taking the limit as \(N\) increases yields the infinite series of the nonnegative numbers \(\int_X f_n\,d\mu\), with value in \([0,\infty]\). Consequently, $$ \int_X \left(\sum_{n=1}^{\infty}f_n\right)\,d\mu = \sum_{n=1}^{\infty}\int_X f_n\,d\mu. $$ This argument also covers the case where the limit is infinite. \(\square\)

The nonnegativity assumption is central. It ensures that the partial sums increase, which is precisely the setting of the monotone convergence theorem. This theorem does not by itself justify the same interchange for a series whose terms can have both signs.

Worked Example: Integrating a Geometric Series

Worked Example: Integrating a Geometric Series

Fix a real number \(r\) with \(0<r<1\), and take \(X=[0,r]\) with Lebesgue measure. For \(n\in\mathbb{N}_0\), let \(f_n(x)=x^n\). These functions are nonnegative and measurable on \([0,r]\). Tonelli’s theorem, with the sequence indexed from \(0\), gives $$ \int_0^r \sum_{n=0}^{\infty}x^n\,dx = \sum_{n=0}^{\infty}\int_0^r x^n\,dx. $$ For every \(x\in[0,r]\), the geometric series has sum \(1/(1-x)\). Hence the left side is $$ \int_0^r \frac{1}{1-x}\,dx = -\log(1-r). $$ On the right, for every \(n\in\mathbb{N}_0\), $$ \int_0^r x^n\,dx = \frac{r^{n+1}}{n+1}. $$ Thus $$ \sum_{n=0}^{\infty}\frac{r^{n+1}}{n+1} = -\log(1-r). $$ For instance, at \(r=\tfrac12\), the integral and the series both equal \(\log 2\). The condition \(r<1\) ensures the geometric-series formula is finite throughout this interval.

The theorem is valid even when the common value is infinite. That possibility is not an exception to the result; it is built into its extended-valued formulation.

Worked Example: An Infinite Value on Both Sides

Take Lebesgue measure on \([0,1]\) and again set \(f_n(x)=x^n\) for \(n\in\mathbb{N}_0\). For \(0\leq x<1\), the pointwise sum is \(1/(1-x)\); at \(x=1\), it is \(+\infty\). The integral of the pointwise sum is infinite, since $$ \int_0^1 \frac{1}{1-x}\,dx=+\infty. $$ On the other hand, each term has integral $$ \int_0^1 x^n\,dx=\frac{1}{n+1}, $$ so the sum of the integrals is the harmonic series $$ \sum_{n=0}^{\infty}\frac{1}{n+1}=+\infty. $$ Tonelli’s theorem gives equality in the extended sense: both sides are \(+\infty\). There is no contradiction with the fact that the pointwise sum is finite for every \(x<1\); its integral can still be infinite.

Finite Integral Means Finiteness Almost Everywhere

The countable-sum identity also gives a useful criterion. If the sum of the individual integrals is finite, then the pointwise sum has a finite integral. A nonnegative measurable function with finite integral can be infinite only on a set of measure zero.

Proposition: Under the hypotheses of Tonelli’s theorem, if $$ \sum_{n=1}^{\infty}\int_X f_n\,d\mu<\infty, $$ then \(\sum_{n=1}^{\infty}f_n(x)<\infty\) for \(\mu\)-almost every \(x\). Moreover, the pointwise sum is integrable and its integral equals the sum of the individual integrals.
Proof. Let \(S=\sum_{n=1}^{\infty}f_n\). Tonelli’s theorem gives $$ \int_X S\,d\mu = \sum_{n=1}^{\infty}\int_X f_n\,d\mu<\infty. $$ If the set \(A=\{x\in X:S(x)=+\infty\}\) had positive measure, then for every \(K>0\) we would have \(S\geq K\mathbf{1}_A\), and therefore $$ \int_X S\,d\mu\geq K\mu(A). $$ If \(\mu(A)=+\infty\), this already contradicts finiteness of the integral. If \(0<\mu(A)<\infty\), the right side grows without bound as \(K\) increases, again contradicting finiteness. Thus \(\mu(A)=0\). The displayed identity also shows that \(S\) is integrable. \(\square\)

“Almost everywhere” matters: the series need not be finite at every point. A divergence at a null set does not prevent the sum from being integrable.

Worked Example: A Series of Indicators on Disjoint Intervals

Let \(X=(0,1)\) with Lebesgue measure. For \(n\in\mathbb{N}\), set $$ I_n=(1-2^{-(n-1)},\,1-2^{-n}] \qquad\text{and}\qquad f_n=\mathbf{1}_{I_n}. $$ The intervals are pairwise disjoint, and their union is \((0,1)\). Each has length \(2^{-n}\), so $$ \int_0^1 f_n(x)\,dx=2^{-n}. $$ Consequently, the sum of the individual integrals is $$ \sum_{n=1}^{\infty}\int_0^1 f_n(x)\,dx = \sum_{n=1}^{\infty}2^{-n} =1. $$ At every \(x\in(0,1)\), exactly one of the indicators equals \(1\), so \(\sum_{n=1}^{\infty}f_n(x)=1\). Its integral is also \(1\). This example shows term-by-term integration for a series of measurable functions and illustrates the finite-integral criterion.

Counting Measure Turns Integrals into Sums

Counting measure on a countable set assigns to each subset its number of elements, with infinite sets having measure \(+\infty\). Integrating a nonnegative function with respect to counting measure is therefore the same as summing its values. Applying Tonelli with counting measure is one way to understand why the order of nonnegative countable sums can be exchanged.

Corollary (nonnegative double sums): If \(a_{ij}\in[0,\infty]\) for \(i,j\in\mathbb{N}\), then $$ \sum_{i=1}^{\infty}\left(\sum_{j=1}^{\infty}a_{ij}\right) = \sum_{j=1}^{\infty}\left(\sum_{i=1}^{\infty}a_{ij}\right). $$ Both sides are understood as extended nonnegative sums.

To see why, equip \(\mathbb{N}\) with counting measure and define \(g_i(j)=a_{ij}\). The countable-sum theorem applied to the sum over \(i\), integrated over the \(j\)-coordinate, says that summing the row sums gives the same result as summing the column sums. Equivalently, both expressions are the integral of the nonnegative function \((i,j)\mapsto a_{ij}\) with respect to counting measure on \(\mathbb{N}\times\mathbb{N}\).

Worked Example: Summing a Nonnegative Matrix in Either Order

Let \(a_{ij}=2^{-i-j}\) for \(i,j\in\mathbb{N}\). For each fixed \(i\), $$ \sum_{j=1}^{\infty}a_{ij} = 2^{-i}\sum_{j=1}^{\infty}2^{-j} = 2^{-i}, $$ because \(\sum_{j=1}^{\infty}2^{-j}=1\). Therefore, $$ \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij} = \sum_{i=1}^{\infty}2^{-i} =1. $$ For each fixed \(j\), the same computation gives \(\sum_{i=1}^{\infty}a_{ij}=2^{-j}\), and hence $$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}a_{ij} = \sum_{j=1}^{\infty}2^{-j} =1. $$ This is a concrete instance of the double-sum corollary. The general corollary does not require the common value to be finite.

Reading the Result Carefully

Tonelli’s countable-sum identity is useful whenever a quantity is expressed as a sum of nonnegative contributions: probabilities of events, indicator functions, or terms in a series of functions. In a probability space, for example, the integral of a nonnegative random variable is its expectation. Thus, if \(A_n\) are measurable events, applying the theorem to \(f_n=\mathbf{1}_{A_n}\) gives $$ \mathbb{E}\!\left[\sum_{n=1}^{\infty}\mathbf{1}_{A_n}\right] = \sum_{n=1}^{\infty}\mathbb{P}(A_n). $$ The random sum on the left counts how many of the events occur, allowing the value \(+\infty\).

A common mistake is to use this formula for a series with positive and negative terms without checking further assumptions. For nonnegative terms, the increasing partial sums supply the required justification, and infinite values are allowed. For signed or complex terms, that monotonicity is unavailable; order changes require additional hypotheses and results.

Check Your Understanding

Use nonnegativity and the monotone convergence theorem to justify each conclusion.

  1. What conditions on \((f_n)\) allow the countable-sum form of Tonelli’s theorem to be applied?
  2. Why does the proof use the partial sums \(S_N=\sum_{n=1}^N f_n\)?
  3. If \(\sum_n\int_X f_n\,d\mu<\infty\), what can be concluded about the pointwise sum, and why is the conclusion only almost everywhere?
  4. For \(a_{ij}\geq0\), why may the two iterated sums be equal even if their common value is infinite?
  5. Which step in the proof would fail to be justified by monotone convergence if the \(f_n\) could take negative values?

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