From Summable Norms to a Convergent Series
A sequence of functions can have small \(L^p\) norms term by term without its pointwise series behaving well. A stronger condition is that the norms themselves have finite sum. For \(1\le p<\infty\), this condition controls not only the \(L^p\) size of the series, but also the sum of the absolute values at almost every point.
Let \((X,\mathcal{M},\mu)\) be any measure space. No assumption of finite measure, sigma-finiteness, or completeness is needed below. Functions may be real-valued or complex-valued, and are taken to be measurable representatives. For \(1\le p<\infty\), \(L^p(\mu)\) consists of equivalence classes modulo equality \(\mu\)-a.e. of measurable functions \(f\) with \(\int |f|^p\,d\mu<\infty\), and \(\|f\|_p=(\int |f|^p\,d\mu)^{1/p}\).
We use Minkowski's inequality in its standard form: if \(u,v\in L^p(\mu)\), then \(\|u+v\|_p\le\|u\|_p+\|v\|_p\). We also use the Monotone Convergence Theorem: for nonnegative measurable \(u_N\uparrow u\), \(\int u_N\,d\mu\uparrow\int u\,d\mu\), allowing infinite values. These results apply on arbitrary measure spaces.
The Summable-Series Theorem
Proof. Put \(C=\sum_{n=1}^{\infty}\|f_n\|_p\), which is finite by hypothesis, and define \(G_N=\sum_{n=1}^{N}|f_n|\). Each \(G_N\) is measurable and nonnegative. Since \(\||f_n|\|_p=\|f_n\|_p\), repeated application of Minkowski's inequality gives $$ \|G_N\|_p\le\sum_{n=1}^{N}\|f_n\|_p\le C. $$ The sequence \(G_N^p\) is nondecreasing and converges pointwise to \(G^p\), where \(G=\sum_{n=1}^{\infty}|f_n|\) is allowed initially to take the value \(+\infty\). The Monotone Convergence Theorem therefore gives $$ \int G^p\,d\mu=\lim_{N\to\infty}\int G_N^p\,d\mu \le C^p. $$ Thus \(G\in L^p(\mu)\), in particular \(G<\infty\) a.e. Indeed, if \(\{G=\infty\}\) had positive measure, then \(\int G^p\,d\mu\) would be infinite. Hence the series \(\sum_n f_n(x)\) converges absolutely outside the measurable null set \(E=\{G=\infty\}\).
Define \(f(x)\) to be the sum of that series on \(E^c\), and set \(f(x)=0\) on \(E\). The partial sums multiplied by \(\chi_{E^c}\) converge pointwise everywhere to \(f\), so \(f\) is measurable. On \(E^c\), \(|f|\le G\), and on \(E\), \(f=0\); consequently \(|f|\le G\) everywhere. It follows that \(\int|f|^p\,d\mu\le\int G^p\,d\mu<\infty\), so \(f\in L^p(\mu)\).
For a fixed \(N\), let \(H_N=\sum_{n>N}|f_n|\). Apply Minkowski's inequality to its finite partial sums and then apply the Monotone Convergence Theorem to their \(p\)-th powers. This yields $$ \|H_N\|_p\le\sum_{n>N}\|f_n\|_p. $$ On \(E^c\), absolute convergence gives \(\left|f-\sum_{n=1}^{N}f_n\right|\le H_N\). This inequality holds a.e., which is sufficient for comparing the integrals defining the \(L^p\) norms. Therefore $$ \left\|f-\sum_{n=1}^{N}f_n\right\|_p \le\|H_N\|_p \le\sum_{n>N}\|f_n\|_p. $$ The right side tends to zero because the series of norms converges. This proves convergence in \(L^p\); the earlier pointwise argument proves absolute convergence a.e. Finally, applying the same finite-sum bound to \(G_N\) and taking the Monotone Convergence limit proves \(\|G\|_p\le C\). \(\square\)
The pointwise majorant \(G=\sum_n|f_n|\) is the central device: it controls every partial sum and the limiting function, while its \(L^p\) norm is bounded by the total sum of the individual norms. The tail version is particularly useful: it gives an explicit error bound using only the norms of the omitted terms.
Worked Examples
Worked Example: A Series on a Counting-Measure Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with counting measure, and for a fixed \(1\le p<\infty\) define \(f_n(k)=3^{-n}\) when \(k=1\), and \(f_n(k)=0\) when \(k\ge2\). Compute the total norm sum and the series limit.
Counting measure gives \(\int h\,d\mu=\sum_{k\ge1}h(k)\) for nonnegative \(h\). Thus $$ \|f_n\|_p=\left(\sum_{k\ge1}|f_n(k)|^p\right)^{1/p} =\left(3^{-np}\right)^{1/p}=3^{-n}. $$ The norm sum is the geometric series $$ \sum_{n=1}^{\infty}\|f_n\|_p =\sum_{n=1}^{\infty}3^{-n} =\frac{1/3}{1-1/3}=\frac12. $$ At \(k=1\), the function series sums to \(\sum_{n\ge1}3^{-n}=\frac12\); at every \(k\ge2\), it sums to zero. Therefore \(f=\frac12\chi_{\{1\}}\) and \(\|f\|_p=(|\frac12|^p)^{1/p}=\frac12\). The majorant \(G\) has exactly the same values, so \(\|G\|_p=\frac12\), matching the total norm sum. Check: substituting \(f(1)=\frac12\) into the counting-measure norm gives \((\sum_k|f(k)|^p)^{1/p}=(2^{-p})^{1/p}=\frac12\).
Worked Example: Disjoint Intervals and a Strict Majorant Bound
On \(\mathbb{R}\) with Lebesgue measure \(m\), let \(p=2\) and \(f_n=2^{-n}\chi_{[n,n+1)}\). Compute \(\sum_n\|f_n\|_2\) and the norm of the sum.
Each interval \([n,n+1)\) has measure \(1\), so $$ \|f_n\|_2 =\left(\int_n^{n+1}4^{-n}\,dx\right)^{1/2} =(4^{-n}\cdot1)^{1/2}=2^{-n}. $$ Hence \(\sum_{n\ge1}\|f_n\|_2=(1/2)/(1-1/2)=1\). The intervals are disjoint, and each \(x\in[1,\infty)\) belongs to exactly one of them. Thus the sum \(f=\sum_n f_n\) equals \(2^{-n}\) on \([n,n+1)\), and is zero outside their union. Its squared norm is $$ \|f\|_2^2=\sum_{n=1}^{\infty}\int_n^{n+1}4^{-n}\,dx =\sum_{n=1}^{\infty}4^{-n} =\frac{1/4}{1-1/4}=\frac13, $$ so \(\|f\|_2=1/\sqrt{3}\). The majorant \(G=\sum_n|f_n|\) equals \(f\) here, so its norm is also \(1/\sqrt{3}\), strictly less than the sum of the norms, \(1\). Check: the theorem's bound reads \(1/\sqrt{3}\le1\), and squaring both positive sides reduces it to \(1/3\le1\). For the tail after \(N\), the squared norm is \(\sum_{n>N}4^{-n}=4^{-N}/3\), so the tail norm is \(2^{-N}/\sqrt{3}\), while the theorem bounds it by \(\sum_{n>N}2^{-n}=2^{-N}\).
Worked Example: Overlapping Terms and the Norm Inequality
On \([0,1)\) with Lebesgue measure, take \(p=2\), \(f_1=\chi_{[0,1)}\), \(f_2=\chi_{[0,1/2)}\), and \(f_n=0\) for \(n\ge3\). Calculate the majorant and compare its norm with the sum of the norms.
The first two norms are $$ \|f_1\|_2=\left(\int_0^1 1\,dx\right)^{1/2}=1,\qquad \|f_2\|_2=\left(\int_0^{1/2}1\,dx\right)^{1/2}=\frac1{\sqrt2}. $$ Thus the total norm sum is \(1+1/\sqrt2\). On \([0,1/2)\), both indicators equal \(1\), so \(G=|f_1|+|f_2|=2\). On \([1/2,1)\), only \(f_1\) is nonzero, so \(G=1\). It follows that $$ \|G\|_2^2=\int_0^{1/2}4\,dx+\int_{1/2}^1 1\,dx =4\cdot\frac12+1\cdot\frac12=\frac52, \qquad \|G\|_2=\sqrt{\frac52}. $$ The bound is strict: \((1+1/\sqrt2)^2=3/2+\sqrt2>5/2\), since \(\sqrt2>1\). Here the series sum is \(f=G\), so this also computes \(\|f\|_2\). Check: direct integration of \(f^2\), which is \(4\) on the first half and \(1\) on the second, again gives \(4(1/2)+1(1/2)=5/2\).
Common Pitfalls and the Role of the Hypothesis
Do not replace the inequality by an equality. In the preceding example, a tempting but incorrect calculation is \(\|f_1+f_2\|_2=\|f_1\|_2+\|f_2\|_2=1+1/\sqrt2\). The actual norm is \(\sqrt{5/2}\), obtained by integrating the square on the two intervals. Minkowski gives an upper bound; equality requires additional structure and is not automatic.
Finite total norm is sufficient, not necessary. For example, on \(\mathbb{R}\) with Lebesgue measure, let \(f_n=\chi_{[n,n+1)}\). Each \(\|f_n\|_p=1\), so the sum of norms is infinite. Nevertheless, at each \(x\) at most one term is nonzero, and the pointwise series converges. This does not contradict the theorem: its hypothesis gives a useful guarantee, not a characterization of every convergent series.
If the hypothesis is dropped, convergence can fail everywhere. On \([0,1]\) with Lebesgue measure, take \(f_n(x)=1\) for every \(n\) and every \(x\). For every finite \(p\), \(\|f_n\|_p=(\int_0^1 1\,dx)^{1/p}=1\), so the norm sum diverges. The partial sums equal \(N\) at every \(x\), and hence the function series does not converge at any point. The theorem therefore concerns a strong and practical sufficient condition: summable \(L^p\) norms produce both a.e. absolute convergence and \(L^p\) convergence.
Check Your Understanding
For each question, identify which part of the theorem controls the claim, and keep a.e. statements distinct from pointwise statements.
- On a counting-measure space, let \(f_n=5^{-n}\chi_{\{2\}}\). Compute \(\sum_n\|f_n\|_p\) and the \(L^p\) norm of the pointwise sum for \(1\le p<\infty\).
- Why does \(\int G^p\,d\mu<\infty\) imply that \(G=\sum_n|f_n|\) is finite a.e.?
- State the tail bound for \(\|f-\sum_{n=1}^N f_n\|_p\), and explain why its right side tends to zero.
- For the two overlapping indicators in the third example, compute \(\|f_1+f_2\|_2^2\) directly from the values on each half of \([0,1)\).
- Does the theorem require \((X,\mathcal{M},\mu)\) to be sigma-finite or complete? Identify where the proof handles the exceptional null set.