Why Partial Sums Are the Right Object
To integrate an infinite series, it is not enough to integrate each term separately and hope that the answers add up. The connection is made by the finite partial sums: they are measurable functions, and when every term is nonnegative they increase pointwise to the series. The Monotone Convergence Theorem then identifies the integral of the limit with the limit of the integrals.
Let \((X,\mathcal{M},\mu)\) be any measure space; it need not be finite, complete, or sigma-finite. Functions in this tutorial take values in \([0,\infty]\). A series of such functions is defined pointwise as the extended-real limit of its partial sums, so its value may be \(+\infty\). The integral identities below are identities of extended nonnegative numbers, and do not assert that either side is finite.
Proof from Increasing Partial Sums
We use the Monotone Convergence Theorem in its standard form: if \(u_N:X\to[0,\infty]\) are measurable and \(u_N\uparrow u\) pointwise on a measure space, then \(u\) is measurable and \(\int u\,d\mu=\lim_N\int u_N\,d\mu\). The proof of that theorem belongs to the earlier material on monotone convergence. We also use finite additivity of the nonnegative Lebesgue integral: for two nonnegative measurable functions \(u,v\), \(\int(u+v)\,d\mu=\int u\,d\mu+\int v\,d\mu\), with \(+\infty\) allowed. Repeated application gives the corresponding identity for a finite sum.
Proof. For each \(N\), \(S_N\) is measurable because it is a finite sum of measurable nonnegative functions. Also \(S_N\le S_{N+1}\) at every point, since \(S_{N+1}=S_N+g_{N+1}\) and \(g_{N+1}\ge0\). By definition, \(S_N(x)\to S(x)\) for each \(x\). Thus the sequence satisfies the measurability, nonnegativity, monotonicity, and pointwise convergence hypotheses of the Monotone Convergence Theorem. In particular, that theorem gives $$ \int_X S\,d\mu=\lim_{N\to\infty}\int_X S_N\,d\mu. $$ Finite additivity gives \(\int_X S_N\,d\mu=\sum_{k=1}^N\int_X g_k\,d\mu\) for every \(N\). Substituting this equality into the preceding limit yields $$ \int_X S\,d\mu =\lim_{N\to\infty}\sum_{k=1}^N\int_X g_k\,d\mu =\sum_{k=1}^{\infty}\int_X g_k\,d\mu. $$ The limit and series are taken in \([0,\infty]\), so this reasoning also covers the case in which the partial sums of integrals are unbounded and the value is \(+\infty\). \(\square\)
The logic is specific: first integrate finite sums, where finite additivity applies; then pass to the infinite sum using monotone convergence. There is no exchange of a limit and an integral without a theorem. In particular, no assumption that the integrals are finite is needed.
Worked Example: Integrating the Exponential Series on a Bounded Interval
On \(([0,1],\mathcal{L},m)\), let \(g_k(x)=x^k/k!\) for integers \(k\ge0\), where \(0!=1\). Each \(g_k\) is a nonnegative measurable function on \([0,1]\). The standard power series for the exponential gives \(\sum_{k=0}^{\infty}g_k(x)=e^x\) at every \(x\in[0,1]\). Apply the theorem, indexing the terms from \(k=0\) in this example:
For each \(k\ge0\), the integral is \(1/k!\) times \([x^{k+1}/(k+1)]_0^1\), hence \(1/(k!(k+1))\). Since \((k+1)!=(k+1)k!\), this is \(1/(k+1)!\). The final series is \(e-1\), by the exponential power series with its \(k=0\) term removed. Check: the antiderivative gives \(\int_0^1e^x\,dx=[e^x]_0^1=e-1\), agreeing with the series calculation. The terms are nonnegative, so the theorem applies without any separate integrability bound.
Worked Example: A Finite Atomic Space with Several Nonzero Terms
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the set of all subsets of \(X\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of points in \(E\). Define three functions by the following values, and set \(g_k=0\) for every \(k\ge4\).
| Point | \(g_1\) | \(g_2\) | \(g_3\) | \(\sum_{k\ge1}g_k\) |
|---|---|---|---|---|
| \(a\) | 1 | 0 | 4 | 5 |
| \(b\) | 0 | 3 | 0 | 3 |
| \(c\) | 2 | 1 | 0 | 3 |
Every function on this finite measurable space is measurable, and all values are nonnegative. The integral with respect to counting measure is the sum of the values at the points. Thus $$ \int_X\sum_{k\ge1}g_k\,d\mu=5+3+3=11. $$ For the right side, calculate each term's integral: \(\int_Xg_1\,d\mu=1+0+2=3\), \(\int_Xg_2\,d\mu=0+3+1=4\), and \(\int_Xg_3\,d\mu=4+0+0=4\). All later integrals are zero, so \(\sum_{k\ge1}\int_Xg_k\,d\mu=3+4+4=11\). Check: summing by points gives \(a:1+0+4=5\), \(b:0+3+0=3\), and \(c:2+1+0=3\), whose total is \(5+3+3=11\), the same value.
Worked Example: The Identity Can Have Infinite Value
Take \(X=\mathbb{R}\) with its Lebesgue sigma-algebra and Lebesgue measure \(m\). For each integer \(k\ge1\), set \(g_k=\chi_{[k,k+1)}\). These are nonnegative measurable functions. The half-open intervals are pairwise disjoint, and their union is \([1,\infty)\). Consequently, their pointwise sum is \(\chi_{[1,\infty)}\), including the endpoint \(1\) and excluding no finite point of that ray.
Each interval \([k,k+1)\) has measure \(1\), so \(\int_{\mathbb{R}}g_k\,dm=1\) for every \(k\), and \(\sum_{k=1}^{\infty}\int_{\mathbb{R}}g_k\,dm=\sum_{k=1}^{\infty}1=+\infty\). The integral of the pointwise sum is \(m([1,\infty))=+\infty\). Check: the \(N\)-th partial sum is \(\chi_{[1,N+1)}\), whose integral is \(m([1,N+1))=N\). These integrals increase without bound, so their limit is \(+\infty\), as the theorem predicts.
Why Measurability and Nonnegativity Matter
Measurability makes the integrals meaningful. The theorem assumes each \(g_k\) is \(\mathcal{M}\)-measurable, so its Lebesgue integral is defined. Measurability also guarantees that the sum is measurable: each \(S_N\) is measurable, and \(S=\lim_NS_N\) is a pointwise limit of measurable functions. More explicitly, for every real \(t\), \(\{S>t\}=\bigcup_{N\ge1}\{S_N>t\}\), a countable union of measurable sets. Without measurability, even a one-term “series” can fail to have a Lebesgue integral: if \(A\subseteq[0,1]\) is not Lebesgue measurable, then \(\chi_A\) is not Lebesgue measurable, because \(\{\chi_A>1/2\}=A\). The theorem cannot be applied to that term.
Nonnegativity makes the partial sums increase. If terms can have either sign, the sequence of partial sums need not increase, and the Monotone Convergence Theorem does not apply. In fact, termwise integration can fail even when each term is integrable and the series converges pointwise. On \(([0,1],\mathcal{L},m)\), define, for \(k\ge1\), $$ f_k=k\chi_{(0,1/k)}-(k+1)\chi_{(0,1/(k+1))}. $$ Each \(f_k\) is measurable and integrable. The two intervals have measures \(1/k\) and \(1/(k+1)\), respectively, so $$ \int_0^1 f_k\,dm=k\cdot\frac1k-(k+1)\cdot\frac1{k+1}=1-1=0. $$ Every finite partial sum telescopes algebraically to $$ \sum_{k=1}^{N}f_k=\chi_{(0,1)}-(N+1)\chi_{(0,1/(N+1))}. $$ For any fixed \(x\in(0,1)\), once \(N+1\ge1/x\), the strict condition \(x<1/(N+1)\) is false, so the second indicator is zero. At \(x=0\) and for \(x\notin(0,1)\), both indicators vanish. Therefore the pointwise series converges everywhere to \(\chi_{(0,1)}\). Its integral is \(1\), whereas the sum of the individual integrals is \(0\). This is a concrete failure of series-integral interchange for signed terms.
The nonnegative theorem needs neither a finite measure space nor a finite value for the resulting integral. Its essential structure is that measurable nonnegative terms give measurable increasing partial sums, with a pointwise limit equal to the series. A useful check before applying it is therefore: are all terms measurable, are all terms nonnegative, and is the series being defined as the pointwise limit of its partial sums?
Check Your Understanding
For each question, identify the measure space and verify the theorem’s hypotheses before using the identity.
- On a measure space \((X,\mathcal{M},\mu)\), suppose \(g_k\ge0\) are measurable. Define their partial sums and explain why they increase pointwise to the series.
- On \([0,1]\) with Lebesgue measure, take \(g_k(x)=x^k\), \(k\ge0\). Compute \(\int_0^1\sum_{k\ge0}g_k\,dm\) using the theorem, allowing the answer \(+\infty\).
- For counting measure on \(X=\{a,b\}\), let \(g_1(a)=2,\ g_1(b)=1,\ g_2(a)=0,\ g_2(b)=4\), with later terms zero. Compute both sides of the interchange identity.
- In the signed example above, compute the integral of the pointwise limit and compare it with \(\sum_k\int f_k\,dm\). Which hypothesis of the theorem fails?
- If a single term is \(\chi_A\) for a non-Lebesgue-measurable set \(A\subseteq[0,1]\), explain why the nonnegative series-integral theorem cannot be applied.