Why Disjointness Simplifies an \(L^p\) Norm
For general functions, expanding \(\left|f+g\right|^p\) does not reduce to a sum of powers. If \(f\) and \(g\) are nonzero on disjoint sets, however, their sum agrees pointwise with whichever function is active there. The integral of its \(p\)th power therefore splits into the two separate integrals. This simple observation gives exact norm calculations without requiring an inequality such as Minkowski’s inequality.
Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space: it is not assumed finite, sigma-finite, or complete. Fix \(p\) with \(1\le p<\infty\). Functions are measurable and real-valued or complex-valued, and statements about their \(L^p\) equivalence classes or norms are understood up to equality \(\mu\)-a.e. For a measurable function \(f\), its nonzero set is \(\{x\in X:f(x)\ne0\}\). We say that \(f_1,\ldots,f_N\) have disjoint supports a.e. when the intersection of the nonzero sets of each distinct pair has measure zero. This definition concerns nonzero sets, not necessarily topological supports.
The measure space need not have finite measure for this definition. The basic integral facts used below are that \(f\mapsto\int f\,d\mu\) is additive for nonnegative measurable functions, including when an integral is infinite, and that \(\int |a\chi_E|^p\,d\mu=|a|^p\mu(E)\) for \(a\ne0\). If \(a=0\), the integral is instead \(0\), regardless of \(\mu(E)\).
The Disjoint-Support Identity
Proof. For each pair \(i\ne j\), let \(N_{ij}=\{f_i\ne0\}\cap\{f_j\ne0\}\). By the hypothesis, \(\mu(N_{ij})=0\). There are only finitely many such pairs, so \(N=\bigcup_{i\ne j}N_{ij}\) is measurable and \(\mu(N)=0\). At every \(x\in X\setminus N\), at most one of \(f_1(x),\ldots,f_N(x)\) is nonzero. If they are all zero, both \(\left|\sum_i f_i(x)\right|^p\) and \(\sum_i|f_i(x)|^p\) equal zero. If just \(f_k(x)\) is nonzero, then \(\sum_i f_i(x)=f_k(x)\), so both quantities equal \(|f_k(x)|^p\). Thus the two nonnegative measurable functions are equal outside the null set \(N\), hence equal a.e. Their integrals are equal. By finite additivity of the integral for nonnegative measurable functions, \(\int\sum_i|f_i|^p\,d\mu=\sum_i\int|f_i|^p\,d\mu\), including when a term is infinite. This proves the identity. If each \(f_i\in L^p(\mu)\), the right side is finite, so the integral on the left is finite and the sum belongs to \(L^p(\mu)\). Taking the \(p\)th root gives the asserted norm formula. \(\square\)
The equality is about the \(p\)th powers of the norms. Taking roots gives \(\left\|\sum_i f_i\right\|_p=(\sum_i\|f_i\|_p^p)^{1/p}\), not generally the sum of the individual norms.
Worked Example: A Sum on Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with its full sigma-algebra and counting measure \(\mu\). This measure space is not finite, though it is sigma-finite. For \(p=2\), take \(f=2\chi_{\{1,2\}}\) and \(g=-3\chi_{\{4\}}\). Their nonzero sets are disjoint, and both functions belong to \(L^2(\mu)\).
Counting measure assigns measure one to each singleton. Therefore $$ \|f\|_2^2=|2|^2\mu(\{1,2\})=4\cdot2=8,\qquad \|g\|_2^2=|-3|^2\mu(\{4\})=9\cdot1=9. $$ The theorem gives \(\|f+g\|_2^2=8+9=17\), hence \(\|f+g\|_2=\sqrt{17}\).
Check: \(f+g\) takes the values \(2,2,-3\) at \(1,2,4\), respectively, and is zero elsewhere. The sum of its squared values is \(2^2+2^2+(-3)^2=4+4+9=17\), agreeing with the theorem.
Indicators and Simple Functions
For indicator functions, the identity turns a norm calculation into a measure calculation. For simple functions, it applies after grouping their nonzero values on disjoint measurable sets. The next corollary also makes explicit how to handle coefficients equal to zero.
Proof. The function \(a_i\chi_{E_i}\) is zero everywhere if \(a_i=0\). For \(a_i\ne0\), its nonzero set is \(E_i\). Thus the nonzero sets of the functions with nonzero coefficients are pairwise disjoint a.e. Apply the theorem to those functions. If there are none, the function on the left is identically zero and both sides are zero. For each retained term, \(\int|a_i\chi_{E_i}|^p\,d\mu=|a_i|^p\mu(E_i)\). This gives the displayed formula. In particular, zero coefficients contribute zero without requiring the undefined product \(0\cdot\infty\). \(\square\)
Worked Example: A Simple Function with Three Values
On \(\mathbb{R}\) with Lebesgue measure \(m\), let $$ h=2\chi_{[0,1)}-\chi_{[1,3)}+3\chi_{[3,4)}. $$ The three intervals are pairwise disjoint, including at their endpoints according to the stated half-open conventions. Their measures are \(1,2,1\), respectively. For \(p=3\), the corollary gives $$ \|h\|_3^3=|2|^3\cdot1+|-1|^3\cdot2+|3|^3\cdot1=8+2+27=37. $$ Thus \(\|h\|_3=\sqrt[3]{37}\).
Check: \(h\) equals \(2\) on \([0,1)\), \(-1\) on \([1,3)\), and \(3\) on \([3,4)\), and is zero outside their union. Integrating \(|h|^3\) directly gives \(8\cdot1+1\cdot2+27\cdot1=37\), as required.
Worked Example: A Zero Coefficient on a Set of Infinite Measure
Take \(X=\mathbb{R}\) with Lebesgue measure, \(p=2\), \(E_1=\mathbb{R}\), \(E_2=[0,1]\), \(a_1=0\), and \(a_2=2\). The sets \(E_1\) and \(E_2\) are not disjoint, but the functions \(a_1\chi_{E_1}=0\) and \(a_2\chi_{E_2}\) have disjoint nonzero sets. The sum is \(2\chi_{[0,1]}\), so $$ \left\|0\chi_{\mathbb{R}}+2\chi_{[0,1]}\right\|_2^2 =|2|^2m([0,1])=4\cdot1=4, $$ and its norm is \(2\). The zero coefficient contributes zero; writing \(|0|^2m(\mathbb{R})\) would produce the undefined expression \(0\cdot\infty\), not a valid term in the formula.
Check: the sum is \(2\) on an interval of measure \(1\) and zero elsewhere, so its squared integral is \(2^2\cdot1=4\).
Worked Example: Disjoint Functions That Are Not Simple
On \(\mathbb{R}\) with Lebesgue measure, define \(f(x)=x\) for \(0<x<1\) and \(f(x)=0\) otherwise, and define \(g(x)=x-2\) for \(2<x<3\) and \(g(x)=0\) otherwise. Their nonzero sets are the disjoint open intervals \((0,1)\) and \((2,3)\). For \(p=2\), direct integration gives $$ \|f\|_2^2=\int_0^1x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13, \qquad \|g\|_2^2=\int_2^3(x-2)^2\,dx =\left[\frac{(x-2)^3}{3}\right]_2^3=\frac13. $$ Consequently \(\|f+g\|_2^2=\frac13+\frac13=\frac23\) and \(\|f+g\|_2=\sqrt{2/3}\).
Check: \(f+g\) equals \(x\) on \((0,1)\), \(x-2\) on \((2,3)\), and zero elsewhere. Its squared integral is \(\int_0^1x^2\,dx+\int_2^3(x-2)^2\,dx=1/3+1/3=2/3\), the same value.
What Fails Without Disjointness
The disjoint-support hypothesis is what makes the pointwise identity in the proof true. For example, on \(\mathbb{R}\) with Lebesgue measure, set \(f=\chi_{[0,1]}\) and \(g=\chi_{[0,1]}\), with \(p=2\). Each function has squared norm \(1\). But \(f+g=2\chi_{[0,1]}\), so \(\|f+g\|_2^2=4\), not \(1+1=2\). The nonzero sets overlap on a set of measure \(1\), and both functions are active there.
Common mistake: For the disjoint functions \(f=2\chi_{\{1,2\}}\) and \(g=-3\chi_{\{4\}}\) in the counting-measure example, one might incorrectly write \(\|f+g\|_2=\|f\|_2+\|g\|_2=\sqrt8+3\). The theorem instead gives $$ \|f+g\|_2=\sqrt{\|f\|_2^2+\|g\|_2^2} =\sqrt{8+9}=\sqrt{17}. $$ The incorrect calculation adds the norms rather than their squares. This distinction matters even when the supports really are disjoint.
Exact additivity of \(p\)th powers is useful whenever a function is built from pieces living on separate measurable regions: indicators, simple functions, and functions defined by different formulas on disjoint sets. It also provides a direct test for membership: under the hypotheses of the theorem, the sum belongs to \(L^p(\mu)\) precisely when the sum of the individual \(p\)th-power integrals is finite. No finiteness assumption on \(\mu(X)\) is needed; what matters is the integral of each active piece.
Check Your Understanding
In each question, specify the measure space and verify whether the nonzero sets are disjoint a.e. before applying the identity.
- For \(p=3\), compute the \(L^3\) norm of \(4\chi_A-2\chi_B\) when \(\mu(A)=2\), \(\mu(B)=5\), and \(\mu(A\cap B)=0\).
- On counting measure on \(\mathbb{N}\), take \(f=3\chi_{\{2,5\}}\) and \(g=4\chi_{\{7\}}\). Find \(\|f+g\|_2\) by the theorem and by summing the squared values directly.
- Explain why the expression \(|0|^p\mu(X)\) should not be used as a term when \(\mu(X)=\infty\), and state the correct contribution of the zero function.
- Give an example of two \(L^2\) functions with overlapping nonzero sets for which the squared norm of their sum is not the sum of their squared norms.
- If a finite family of functions has pairwise disjoint nonzero sets a.e. and each belongs to \(L^p(\mu)\), explain why their sum also belongs to \(L^p(\mu)\).