Countable Sums as Increasing Limits
Integrating a series term by term is useful whenever a nonnegative function is presented as a sum of simpler measurable functions. The key is that the partial sums increase to the full sum. Monotone Convergence then gives an identity even when the sum or integral is infinite; neither finiteness of the measure space nor integrability of the individual terms is needed.
Let \((X,\mathcal{M},\mu)\) be any measure space, with no assumption that \(\mu\) is finite or sigma-finite. In this tutorial all functions in countable sums take values in \([0,\infty]\) and are measurable. The sum is defined pointwise as the increasing limit of its finite partial sums, so it may take the value \(+\infty\).
Proof. For \(N\ge1\), define the partial sum \(S_N=\sum_{k=1}^{N}g_k\). A finite sum of nonnegative measurable functions is measurable, and \(S_N\le S_{N+1}\) at every point. Define \(S=\sup_{N\ge1}S_N\). A pointwise supremum of measurable functions is measurable, and by the definition of a nonnegative countable sum, \(S=\sum_{k=1}^{\infty}g_k\). Monotone Convergence applies to the nonnegative measurable sequence \((S_N)\), which increases pointwise everywhere to \(S\). Therefore \(\int_X S\,d\mu=\lim_{N\to\infty}\int_X S_N\,d\mu\). Additivity of the integral for finite sums gives \(\int_X S_N\,d\mu=\sum_{k=1}^{N}\int_Xg_k\,d\mu\). These finite sums increase to the extended series on the right, proving the identity. \(\square\)
The hypotheses are precisely nonnegativity and measurability of every term. The measure space itself need not be sigma-finite for this one-space identity. If the terms are signed or complex-valued, this theorem does not apply: cancellation can make the pointwise sum behave differently from the sum of the integrals.
How Counting Measure Turns Sums into Integrals
For a countable set \(I\), counting measure assigns to each subset its number of elements, with value \(+\infty\) for an infinite set. Integration with respect to counting measure is summation. Thus a countable family of integrals can itself be viewed as an integral over the index set.
Proof. For any real \(a\), the set \(\{(k,x):G(k,x)>a\}\) is \(\bigcup_{k=1}^{\infty}\bigl(\{k\}\times\{x:g_k(x)>a\}\bigr)\). Each factor set is measurable, so this countable union belongs to the product sigma-algebra. Hence \(G\) is measurable. Counting measure on \(\mathbb{N}\) is sigma-finite, since \(\mathbb{N}=\bigcup_{k\ge1}\{k\}\) and each singleton has measure one; \(\mu\) is sigma-finite by hypothesis. The nonnegative, measurable version of Tonelli’s theorem therefore applies to \(G\). Integrating first over the counting-measure coordinate gives \(\int_X\sum_k g_k\,d\mu\), while integrating first over \(X\) gives \(\sum_k\int_Xg_k\,d\mu\). These are the asserted equalities. \(\square\)
This product-space viewpoint explains the phrase “under Tonelli.” The preceding theorem is valid without sigma-finiteness because it follows directly from Monotone Convergence. The product-measure argument above uses the stated sigma-finiteness assumptions to invoke Tonelli.
Worked Example: A Double Series with Counting Measure
On \(\mathbb{N}\) with counting measure \(\#\), let \(g_k(n)=2^{-k}3^{-n}\) for \(k,n\ge1\). Compute the sum over \(k\) of the integrals, and verify it by summing in the other order.
Every \(g_k\) is nonnegative and measurable because every subset of \(\mathbb{N}\) is measurable. Integration with respect to counting measure is summation. First, the geometric series over \(n\) is \(\sum_{n=1}^{\infty}3^{-n}=(1/3)/(1-1/3)=1/2\). Thus, for each fixed \(k\),
Summing over \(k\), and using \(\sum_{k=1}^{\infty}2^{-k}=(1/2)/(1-1/2)=1\), gives \(\sum_{k=1}^{\infty}\int_{\mathbb{N}}g_k\,d\#=\frac12\cdot1=\frac12\). In the reverse order, for fixed \(n\), \(\sum_{k=1}^{\infty}2^{-k}3^{-n}=3^{-n}\cdot1=3^{-n}\), and summing over \(n\) gives \(1/2\). Tonelli applies because both factors, counting measure on \(\mathbb{N}\) and counting measure on \(\mathbb{N}\), are sigma-finite and the array is nonnegative. Check: both orders evaluate to \((1/2)(1)=1/2\).
Exact Lebesgue-Measure Examples
Worked Example: A Series of Indicators on the Half-Line
On \((\mathbb{R},\mathcal{L},m)\), define \(g_k(x)=2^{-k}\chi_{[0,k]}(x)\) for \(k\ge1\). Compute the integral of the pointwise sum and compare it with the sum of the individual integrals.
Each \(g_k\) is nonnegative and Lebesgue measurable. Since \(m([0,k])=k\), its integral is \(\int_{\mathbb{R}}g_k\,dm=2^{-k}k\). The finite-sum formula, for \(r\ne1\), is \(\sum_{k=1}^{N}kr^k=r\bigl(1-(N+1)r^N+Nr^{N+1}\bigr)/(1-r)^2\). Substituting \(r=1/2\) gives \(\sum_{k=1}^{N}k2^{-k}=2-(N+2)/2^N\), which tends to \(2\). Therefore \(\sum_{k=1}^{\infty}\int_{\mathbb{R}}g_k\,dm=2\).
For a direct check of the integral of the sum, consider \(x\in(j,j+1)\), where \(j\) is any integer with \(j\ge0\). The condition \(x\in[0,k]\) holds exactly when \(k\ge j+1\). Hence $$ \sum_{k=1}^{\infty}g_k(x)=\sum_{k=j+1}^{\infty}2^{-k}=2^{-j}. $$ The endpoints of these intervals form a countable set and have Lebesgue measure zero. Integrating the resulting function over the intervals gives \(\sum_{j=0}^{\infty}2^{-j}\cdot m((j,j+1))=\sum_{j=0}^{\infty}2^{-j}=2\). At \(x=0\), the sum is also \(\sum_{k\ge1}2^{-k}=1\), which does not change the integral. The direct calculation agrees with the countable-sum theorem.
Worked Example: Disjoint Supports and a Finite Total
On \((\mathbb{R},\mathcal{L},m)\), define \(h_k(x)=2^{-k}\chi_{[k,k+1)}(x)\) for \(k\ge1\). Find the pointwise sum and its integral.
The intervals \([k,k+1)\) are pairwise disjoint: if \(k<\ell\), then \(k+1\le\ell\), and the point \(k+1\) is excluded from \([k,k+1)\). Each interval has Lebesgue measure \(1\), so \(\int_{\mathbb{R}}h_k\,dm=2^{-k}\cdot1=2^{-k}\). Summing the integrals gives \(\sum_{k=1}^{\infty}2^{-k}=(1/2)/(1-1/2)=1\).
At each \(x\), at most one indicator is nonzero, so \(\sum_{k=1}^{\infty}h_k(x)\) equals \(2^{-k}\) if \(x\in[k,k+1)\) for some \(k\ge1\), and equals \(0\) otherwise. Its integral is therefore \(\sum_{k=1}^{\infty}2^{-k}m([k,k+1))=\sum_{k=1}^{\infty}2^{-k}=1\). This also checks the theorem directly from the disjoint pieces, including the points at the interval endpoints.
Why Nonnegativity Cannot Be Dropped
A common mistake is to use the countable-sum identity for signed functions merely because every individual term has a finite integral. The identity proved above depends on increasing nonnegative partial sums; signed partial sums need not increase, and cancellation can invalidate the conclusion.
Worked Example: A Signed Series That Does Not Interchange
On \(([1,\infty),\mathcal{L},m)\), define \(f_n=\chi_{[n,n+1)}-\chi_{[n+1,n+2)}\), for \(n\ge1\). Compare the integral of the pointwise sum with the sum of the integrals.
For each \(n\), both intervals have measure \(1\), so \(\int f_n\,dm=1-1=0\). Consequently \(\sum_{n=1}^{\infty}\int f_n\,dm=0\). To find the pointwise series, first take \(x\in[1,2)\): only the positive term for \(n=1\) contributes, so the sum is \(1\). If \(x\in[k,k+1)\) for an integer \(k\ge2\), the negative term for \(n=k-1\) contributes \(-1\), and the positive term for \(n=k\) contributes \(1\); all other terms vanish. Thus the pointwise sum is \(\chi_{[1,2)}\), whose integral is \(1\cdot m([1,2))=1\). The two values, \(1\) and \(0\), differ. Each pointwise series converges, but the terms are not nonnegative, so the theorem does not apply.
For nonnegative measurable functions, countable summation and integration are compatible because finite partial sums increase to the full sum. In applications, check those two conditions before interchanging the operations. If signs or complex values are present, a separate theorem with appropriate integrability hypotheses is needed.
Check Your Understanding
For each question, identify the measure space and verify the relevant hypotheses before calculating.
- On \(\mathbb{N}\) with counting measure, let \(g_k(n)=4^{-k}5^{-n}\). Compute \(\sum_k\int g_k\,d\#\), and verify the value by summing in the other order.
- On \(\mathbb{R}\) with Lebesgue measure, let \(g_k=3^{-k}\chi_{[0,k]}\). Compute \(\sum_k\int g_k\,dm\).
- Explain why the countable-sum theorem applies to \(g_k=2^{-k}\chi_{[k,k+1)}\), and identify the pointwise sum.
- In the signed-series example, calculate the pointwise sum separately on \([1,2)\) and \([k,k+1)\) for each integer \(k\ge2\).
- Does the countable-sum identity require \(\mu(X)<\infty\)? State which condition on the functions is essential in the theorem.