Two Different Questions About Partial Sums
For a series of nonnegative measurable functions, there are two natural ways to ask whether its partial sums remain bounded. At a fixed point \(x\), one can ask whether the numerical partial sums stay bounded as the number of terms increases. Separately, one can ask whether the integrals of those partial sums stay bounded. These questions are related, but they are not equivalent pointwise: an integral can be finite even when the series diverges at points in a null set, and the integrals can diverge even when the series is finite at every point.
Throughout the general statements, \((X,\mathcal{M},\mu)\) is an arbitrary measure space, with no assumption that \(\mu\) is finite, sigma-finite, or complete. The functions \(g_k:X\to[0,\infty]\) are measurable. Define the finite partial sums \(S_N=\sum_{k=1}^{N}g_k\), and define the infinite sum \(S=\sum_{k=1}^{\infty}g_k=\lim_{N\to\infty}S_N\), allowing the value \(+\infty\). The sequence \(S_N\) is nondecreasing at every point. Its pointwise limit \(S\) is measurable.
The finite-sum identity holds for nonnegative measurable functions, with infinite values allowed. The Monotone Convergence Theorem, established earlier, applies to \(S_N\uparrow S\) and gives \(I_N\uparrow I\). Thus the integrals of the finite partial sums never decrease, and their limit is exactly the integral of the infinite series. This statement alone does not say whether that common limit is finite or infinite.
When Are the Partial-Sum Integrals Bounded?
- \(\sup_N\int_X S_N\,d\mu<\infty\);
- \(\int_X S\,d\mu<\infty\).
Proof. By the Monotone Convergence Theorem, \(\int_X S\,d\mu=\lim_N\int_X S_N\,d\mu=\sup_N\int_X S_N\,d\mu\), where the equality is in \([0,\infty]\). Therefore the supremum is finite if and only if \(\int_XS\,d\mu\) is finite. Suppose its value is \(C<\infty\). For each positive integer \(j\), let \(E=\{x:S(x)=\infty\}\). Since \(S\ge j\chi_E\), monotonicity of the nonnegative integral gives \(j\mu(E)\le\int_XS\,d\mu=C\). Hence \(\mu(E)\le C/j\) for every \(j\), which forces \(\mu(E)=0\). Thus \(S<\infty\) \(\mu\)-a.e. Finally, if the partial-sum integrals are unbounded, their nondecreasing limit is \(+\infty\), and the same theorem gives \(\int_XS\,d\mu=+\infty\). \(\square\)
The proposition concerns boundedness of the integrals, not a uniform bound on \(S_N(x)\) over all \(x\). It also does not say that finite integral is equivalent to pointwise convergence of the series. For nonnegative terms, \(S_N(x)\) is bounded in \(N\) exactly when \(S(x)<\infty\); the proposition only concludes this pointwise finiteness almost everywhere when the integral is finite.
To verify the consequence, if \(\mu(\{S=\infty\})>0\), then for every positive integer \(j\), \(S\ge j\chi_{\{S=\infty\}}\), so \(\int_XS\,d\mu\ge j\mu(\{S=\infty\})\). If that set has infinite measure, the integral is already infinite; if it has positive finite measure, the lower bounds grow without bound as \(j\to\infty\).
Worked Examples
Worked Example: A Geometric Series with Bounded Integrals
On \(([0,1],\mathcal{L},m)\), where \(m\) is Lebesgue measure, let \(g_k(x)=2^{-k}\) for every \(x\in[0,1]\) and \(k\ge1\). Compute the finite partial-sum integrals and their limit.
For every \(x\in[0,1]\), the geometric sum gives \(S_N(x)=\sum_{k=1}^N2^{-k}=1-2^{-N}\). Since \(m([0,1])=1\),
These integrals increase and are bounded above by \(1\); their limit is \(1\). Pointwise, \(S_N(x)\to1\) for every \(x\), so \(S(x)=1\) and \(\int_{[0,1]}S\,dm=1\). Check: at \(N=1\), the integral is \(1-2^{-1}=1/2\); at \(N=2\), it is \(1-2^{-2}=3/4\). Both values agree with integrating the constant partial sums directly over an interval of measure \(1\).
Worked Example: Diverging Integrals but Finite Pointwise Sums
On \(((0,1],\mathcal{L},m)\), set \(g_n=\chi_{(0,1/n)}\) for \(n\ge1\). Compute the integrals of the partial sums and compare them with the pointwise series.
The interval \((0,1/n)\) lies in \((0,1]\) and has measure \(1/n\). Finite additivity of the integral for nonnegative functions gives
These harmonic partial sums diverge. In detail, for each integer \(r\ge1\), the terms with \(2^{j-1}<n\le2^j\), for \(j=1,\ldots,r\), contribute at least \(2^{j-1}\cdot2^{-j}=1/2\) in each block. Including the \(n=1\) term gives \(H_{2^r}\ge1+r/2\), which is unbounded as \(r\to\infty\).
At a fixed \(x\in(0,1]\), the \(n\)-th term is \(1\) precisely when \(0<x<1/n\), or equivalently \(n<1/x\). There are only finitely many positive integers satisfying this strict inequality for each \(x>0\). Thus \(S(x)<\infty\) at every point of \((0,1]\), even though \(\int_{(0,1]}S\,dm=+\infty\) by Monotone Convergence. Check: at \(x=1\), no positive integer satisfies \(n<1\), so \(S(1)=0\); at \(x=1/4\), precisely \(n=1,2,3\) contribute, so \(S(1/4)=3\). The series is pointwise finite, but its integral is infinite.
Worked Example: Pointwise Divergence on a Null Set
On \((\mathbb{R},\mathcal{L},m)\), let \(g_n=\chi_{\{0\}}\) for every \(n\ge1\). Find the pointwise partial sums and their integrals.
At \(x=0\), each summand equals \(1\), so \(S_N(0)=N\) and \(S(0)=+\infty\). At every \(x\ne0\), each summand equals \(0\), so \(S_N(x)=S(x)=0\). Since the singleton \(\{0\}\) has Lebesgue measure zero,
Thus the integral totals remain bounded although the series diverges at one point. Check: for \(N=5\), \(S_5(0)=5\), but its integral is \(5\cdot0=0\). The proposition is consistent with this: \(S<\infty\) almost everywhere, because its infinite-value set \(\{0\}\) is null.
What Each Type of Divergence Tells You
For a nonnegative measurable series, the following conclusions should be kept separate:
| Information about the partial sums | What follows | What does not follow |
|---|---|---|
| \(\sup_N\int_XS_N\,d\mu<\infty\) | \(\int_XS\,d\mu<\infty\) and \(S<\infty\) \(\mu\)-a.e. | It does not imply \(S_N(x)\) is bounded at every point. |
| \(\int_XS_N\,d\mu\to+\infty\) | \(\int_XS\,d\mu=+\infty\). | It does not imply \(S(x)=+\infty\) on a set of positive measure. |
| \(S_N(x)\to+\infty\) on a set of positive measure | \(\int_XS\,d\mu=+\infty\), hence the integral totals are unbounded. | It does not follow if divergence occurs only on a null set. |
A common mistake is to infer pointwise divergence from unbounded integral totals. The calculation in the indicator example shows why this is wrong: \(\int S_N\,dm=H_N\to\infty\), but for every \(x\in(0,1]\), only finitely many terms are nonzero. The correction is to conclude only that the integral of the limiting sum is infinite. Conversely, the singleton example shows that pointwise divergence at even one point does not force the integral totals to grow when that point is null.
These distinctions are useful whenever a series is integrated term by term. The increasing partial sums determine the integral of the full series through Monotone Convergence, including when that value is \(+\infty\). Bounded partial-sum integrals provide exactly the additional information that the series has finite integral and is finite almost everywhere; they do not provide pointwise convergence everywhere or a uniform pointwise bound.
Check Your Understanding
For each question, distinguish the behavior of the pointwise sums from the behavior of their integrals.
- On \([0,1]\) with Lebesgue measure, let \(g_n=3^{-n}\). Compute \(\int S_N\,dm\) and its limit.
- For \(g_n=\chi_{(0,1/n)}\) on \((0,1]\), explain why \(S(x)\) is finite for every \(x>0\) although \(\int S\,dm=+\infty\).
- Give a reason that bounded \(\int S_N\,d\mu\) implies \(S<\infty\) \(\mu\)-a.e., but need not imply finiteness at every point.
- If \(S=+\infty\) on a set of positive measure, show that \(\int S\,d\mu=+\infty\).
- On Lebesgue measure on \(\mathbb{R}\), calculate \(S_N(0)\) and \(\int S_N\,dm\) when every term is \(\chi_{\{0\}}\).