Disjoint Supports Turn a Norm into a Sum
For general functions, the norm of a sum depends on how the functions overlap and on cancellation between them. Pairwise disjoint supports remove both complications: at almost every point, at most one summand is nonzero. For \(1\le p<\infty\), this makes the \(p\)-th power of the norm additive across the summands. The resulting identity gives an exact criterion for when a series of such functions converges in \(L^p\).
Throughout, \((X,\mathcal{M},\mu)\) is any measure space; no finiteness, completeness, or sigma-finiteness assumption is needed. Fix \(1\le p<\infty\). Functions may be real-valued or complex-valued and are measurable. The space \(L^p(\mu)\) consists of equivalence classes, under equality \(\mu\)-a.e., of functions \(f\) with \(\int_X |f|^p\,d\mu<\infty\), and \(\|f\|_p=(\int_X|f|^p\,d\mu)^{1/p}\). For a chosen representative, its nonzero set is \(\{x:f(x)\ne0\}\); here, saying that functions have pairwise disjoint supports means that the intersections of their nonzero sets have measure zero.
Proof. Choose measurable representatives and let \(E\) be the union of all intersections of two distinct nonzero sets among these finitely many representatives. Each such intersection is null by hypothesis, so \(E\in\mathcal{M}\) and \(\mu(E)=0\). Outside \(E\), at most one representative is nonzero. Therefore, at every \(x\notin E\), \(\left|\sum_{k=1}^N f_k(x)\right|^p=\sum_{k=1}^N|f_k(x)|^p\). The equality holds \(\mu\)-a.e.; integrating both sides and using finite additivity of the integral for nonnegative integrable functions gives the stated identity. Changing values on \(E\) does not change any of the \(L^p\) classes or norms. \(\square\)
The Convergence Criterion
Proof. First, the finite-sum identity gives, for \(M>N\), \(\|S_M-S_N\|_p^p=\sum_{k=N+1}^M\|f_k\|_p^p\). If \(S_N\) converges in \(L^p(\mu)\), it is Cauchy in that norm. Taking the Cauchy condition with tolerance \(1\), there is an \(N_0\) such that \(\sum_{k=N+1}^M\|f_k\|_p^p<1\) whenever \(M>N\ge N_0\). In particular, the partial sums of the nonnegative numerical series are bounded: the terms through \(N_0\) have a finite sum, and every later partial tail is less than \(1\). Hence \(\sum_k\|f_k\|_p^p<\infty\).
Conversely, suppose \(\sum_k\|f_k\|_p^p<\infty\). Choose measurable representatives, let \(A_k=\{x:f_k(x)\ne0\}\), and set \(E=\bigcup_{i<j}(A_i\cap A_j)\). This is a countable union of measurable null sets, so \(E\) is measurable and null. Define \(g_k=f_k\chi_{X\setminus E}\). Each \(g_k\) represents the same \(L^p\) class as \(f_k\), and at every point of \(X\), at most one \(g_k\) is nonzero. This modification is important: disjointness was assumed only almost everywhere, so the original representatives need not have pointwise disjoint nonzero sets on \(E\).
Put \(T_N=\sum_{k=1}^N g_k\). At each point, these partial sums eventually become constant: if one \(g_j(x)\ne0\), then \(T_N(x)=g_j(x)\) for all \(N\ge j\); if all \(g_j(x)=0\), then \(T_N(x)=0\) for every \(N\). Thus \(T_N\) has an everywhere-defined pointwise limit \(S\), which is measurable as a pointwise limit of measurable functions. Also, \(|T_N|^p\) increases pointwise to \(|S|^p\): at each point it is either always zero or changes from zero to one fixed nonnegative value and then stays there.
The Monotone Convergence Theorem applies to the nonnegative measurable functions \(|T_N|^p\), which increase at every point. By the finite-sum identity and \(g_k=f_k\) a.e., \(\int_X|T_N|^p\,d\mu=\sum_{k=1}^N\|f_k\|_p^p\). Therefore \(\int_X|S|^p\,d\mu=\lim_N\sum_{k=1}^N\|f_k\|_p^p<\infty\), so \(S\in L^p(\mu)\), and this also proves the asserted formula for \(\|S\|_p^p\).
For the tail, at every point \(S-T_N\) is zero or equals the unique later nonzero \(g_k\). Consequently, \(|S-T_N|^p=\lim_{M\to\infty}\left|\sum_{k=N+1}^M g_k\right|^p\), with the finite tail powers increasing pointwise. Applying the Monotone Convergence Theorem again, and then the finite-sum identity to each finite tail, gives \(\|S-T_N\|_p^p=\sum_{k=N+1}^{\infty}\|f_k\|_p^p\). The numerical series converges, so its tails tend to zero. Since \(T_N=S_N\) as \(L^p\) classes, \(S_N\to S\) in \(L^p(\mu)\), completing both directions. \(\square\)
Worked Consequences
Worked Example: A Convergent Series on the Real Line
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. Fix \(1\le p<\infty\), and define \(f_k=2^{-k/p}\chi_{[k-1,k)}\) for \(k\ge1\). The intervals are pairwise disjoint, including their endpoint convention: each right endpoint is excluded and belongs to the next interval.
Each interval has measure \(1\), and \((2^{-k/p})^p=2^{-k}\). Thus \(\|f_k\|_p^p=\int_{\mathbb{R}}|f_k|^p\,dm=2^{-k}\cdot1=2^{-k}\). The geometric series has sum \(\sum_{k=1}^\infty2^{-k}=(1/2)/(1-1/2)=1\). The theorem gives an \(L^p\) limit \(S=\sum_{k\ge1}f_k\) and \(\|S\|_p^p=1\), so \(\|S\|_p=1\).
For a direct tail check, \(\|S-S_N\|_p^p=\sum_{k=N+1}^\infty2^{-k}=2^{-N}\), hence \(\|S-S_N\|_p=2^{-N/p}\). At \(N=2\), the tail’s \(p\)-th power is \(2^{-3}+2^{-4}+\cdots=(1/8)/(1-1/2)=1/4=2^{-2}\), as the formula predicts. Check: on each interval \([k-1,k)\), \(S\) equals \(2^{-k/p}\); summing its \(p\)-th power integral gives \(\sum_{k\ge1}2^{-k}=1\), the same norm calculation.
Worked Example: Disjoint Supports Do Not Ensure Convergence
On the same Lebesgue measure space, fix \(1\le p<\infty\) and let \(h_k=k^{-1/p}\chi_{[k-1,k)}\). The supports are pairwise disjoint, and each interval again has measure \(1\). Therefore \(\|h_k\|_p^p=(k^{-1/p})^p\cdot1=1/k\), and the finite-sum identity gives \(\|\sum_{k=1}^N h_k\|_p^p=H_N=\sum_{k=1}^N1/k\).
The harmonic series diverges: for each integer \(j\ge1\), the terms with \(2^{j-1}<k\le2^j\) number \(2^{j-1}\), and each is at least \(1/2^j\). That block sums to at least \(2^{j-1}/2^j=1/2\). Infinitely many such blocks force \(H_N\) to be unbounded. Hence \(\sum_k\|h_k\|_p^p=\infty\), and the criterion says the partial sums do not converge in \(L^p\). For example, \(H_3=1+1/2+1/3=11/6\), so the third partial sum has norm \((11/6)^{1/p}\); these partial-sum norms cannot remain bounded as \(N\) grows. Check: the exact formula \(\|S_N\|_p^p=H_N\) agrees with integrating separately on the \(N\) disjoint unit intervals.
Worked Example: Disjointness Almost Everywhere and a Null Set
Let \(X=\mathbb{R}\) with Lebesgue measure and, for every \(k\ge1\), take \(q_k=\chi_{\{0\}}\). The nonzero set of every representative is \(\{0\}\), so the supports are not disjoint pointwise; nevertheless, every pairwise intersection is the null set \(\{0\}\). Each norm satisfies \(\|q_k\|_p^p=\int_{\{0\}}1\,dm=0\), and the sum of the component norm powers is \(0\).
The raw partial sums satisfy \(\sum_{k=1}^Nq_k(0)=N\), while they are zero at every \(x\ne0\). To apply the theorem pointwise, the common exceptional set is \(E=\{0\}\). Replacing each representative by \(r_k=q_k\chi_{\mathbb{R}\setminus E}\) gives \(r_k=0\) everywhere. Their partial sums and limit are identically zero, and their \(L^p\) norms are all zero. Check: the resulting limit has \(\|0\|_p^p=0=\sum_k\|q_k\|_p^p\), exactly as the theorem states. The original representatives need not converge pointwise to a function chosen to be zero on \(E\); the modification ensures the pointwise-limit argument is valid everywhere.
A Common Mistake and the Scope of the Identity
Common mistake: treating almost-everywhere disjointness as pointwise disjointness. In the last example, take \(p=2\) and \(N=2\). A pointwise calculation at \(0\) gives \(|q_1(0)+q_2(0)|^2=|2|^2=4\), whereas \(|q_1(0)|^2+|q_2(0)|^2=1+1=2\). So the pointwise identity is false there. The correction is to remove the common null set first, or to assert the identity only almost everywhere and then integrate. In this example, both integrals are zero because the discrepancy occurs only at \(\{0\}\).
The criterion is special to finite \(p\): its central quantity is the sum of the \(p\)-th powers of the component norms. It should not be transferred unchanged to \(L^\infty(\mu)\), whose norm is the essential supremum rather than an integral of a finite power. Nor does the identity hold for arbitrary overlapping supports: for example, two copies of \(\chi_{[0,1)}\) on \(\mathbb{R}\) with Lebesgue measure have sum with squared \(L^2\) norm \(4\), while the sum of their squared norms is \(1+1=2\). Disjointness is precisely what removes that overlap.
Check Your Understanding
Use the finite-sum identity and convergence criterion, and state where disjointness is being used.
- For \(f_k=3^{-k/p}\chi_{[k-1,k)}\) on \(\mathbb{R}\), compute \(\sum_k\|f_k\|_p^p\) and the norm of the series limit.
- If pairwise supports are disjoint \(\mu\)-a.e., why is it useful to remove the union of all pairwise intersections before constructing a pointwise series limit?
- For \(g_k=k^{-1/2}\chi_{[k-1,k)}\) on \(\mathbb{R}\), determine whether the partial sums converge in \(L^2\), and justify the conclusion.
- State the exact formula for \(\|S-S_N\|_p^p\) when the disjoint-support series converges in \(L^p(\mu)\).