Why the Total Measure Matters
For a constant function, every point of the space contributes the same amount to the integral of its \(p\)th power. If the space has infinite measure, that repeated contribution can make the integral infinite. An indicator function behaves differently: only the measure of the set where it equals one matters. These two basic cases distinguish a global property of the measure space from a local property of a function’s support.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space; it need not be complete or sigma-finite. The functions considered are measurable and real-valued or complex-valued. Fix \(1\le p<\infty\). The space \(L^p(\mu)\) consists of equivalence classes, up to equality \(\mu\)-a.e., of functions \(f\) for which \(\int_X |f|^p\,d\mu<\infty\); their norm is \(\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}\). We will also say that the extended integral defining this expression is infinite when the function is not in \(L^p(\mu)\). That infinite value is not an \(L^p\) norm.
Constants and Indicators
Proof. If \(c=0\), then \(|f|^p=0\) everywhere, so its integral is zero regardless of \(\mu(X)\). If \(c\ne0\), then \(|f|^p\) is the nonnegative constant \(|c|^p\). By the integral of a nonnegative simple function, its integral is \(|c|^p\mu(X)\); if \(\mu(X)=+\infty\), this is \(+\infty\). Taking the \(p\)th root when the integral is finite gives \(\|f\|_p=|c|\mu(X)^{1/p}\), including the case \(\mu(X)=0\). For the indicator, \(|\chi_E|^p=\chi_E\) at every point, since its values are zero and one. Its integral is therefore \(\mu(E)\), which is finite exactly when \(\chi_E\in L^p(\mu)\). Taking a \(p\)th root gives the stated norm. \(\square\)
The zero constant is an important exception: it belongs to \(L^p(\mu)\) even when the whole space has infinite measure. Likewise, a set \(E\) of measure zero gives \(\|\chi_E\|_p=0\), so its indicator represents the zero element of \(L^p(\mu)\), even if \(E\) is not empty. Membership is determined by measure, not by whether a set contains points.
Worked Example: Constants and Indicators on a Finite Interval
Take \(X=[0,4]\) with the Lebesgue sigma-algebra and Lebesgue measure \(m\). Let \(f(x)=-2\), let \(E=[1,3)\), and use \(p=2\). The interval \(X\) has measure \(4\), and \(E\) has measure \(3-1=2\). For the constant, \(|f(x)|^2=|-2|^2=4\), so
For the indicator, \(|\chi_E|^2=\chi_E\), so
Check: integrating directly, \(\int_0^4|-2|^2\,dx=\int_0^4 4\,dx=4\cdot4=16\), and \(\int_0^4\chi_{[1,3)}(x)\,dx=\int_1^3 1\,dx=3-1=2\). These reproduce the two integrals used to compute the norms.
Worked Example: An Infinite Space with a Finite-Measure Set
Take \(X=\mathbb{R}\), its Lebesgue sigma-algebra, and Lebesgue measure \(m\). The space has infinite measure. For the constant \(f(x)=1\), with \(p=2\), the integral over all of \(\mathbb{R}\) is infinite. Indeed, on each interval \([-N,N]\), where \(N\) is a positive integer,
As \(N\) grows, these lower bounds are unbounded, so \(\int_{\mathbb{R}}|f|^2\,dm=+\infty\). Thus \(f\notin L^2(m)\). In contrast, for \(E=[-1,2]\), the length is \(2-(-1)=3\), and
For \(H=[0,\infty)\), every interval \([0,N]\) is contained in \(H\) and has measure \(N\). Hence \(m(H)\ge N\) for every positive integer \(N\), so \(m(H)=+\infty\) and \(\chi_H\notin L^2(m)\). Check: the integral of \(|\chi_H|^2\) over \([0,N]\) is \(N\), while the integral of \(|\chi_E|^2\) over \(\mathbb{R}\) is \(3\). The first grows without bound; the second is the length of its finite interval.
Worked Example: Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be all subsets of \(\mathbb{N}\), and let \(\mu\) be counting measure. Thus the integral of a nonnegative function is the sum of its values, and \(\mu(\mathbb{N})=+\infty\). For \(f(n)=3\) and \(p=2\), the partial sums of the integral are
These sums are unbounded as \(N\to\infty\), so \(\int_{\mathbb{N}}|f|^2\,d\mu=+\infty\), and this nonzero constant is not in \(L^2(\mu)\). Now let \(F=\{2,5,9\}\). It has exactly three elements, so
For comparison, the even integers form an infinite set: their first \(N\) elements contribute \(N\) to the integral of their indicator’s square. Thus that indicator has infinite integral and is not in \(L^2(\mu)\). Check: the three nonzero values for \(\chi_F\) give integral \(3\), while the partial sums for the even-integer indicator are \(1,2,\ldots,N\) and are unbounded.
A Consequence on Finite-Measure Spaces
The same finite-versus-infinite distinction affects comparisons between different finite exponents. On a finite-measure space, integrability at a larger exponent implies integrability at a smaller one. This consequence is useful, but it does not say that every function in the smaller space belongs to the larger space.
Proof. Choose a measurable representative of \(f\), and partition \(X\) into \(A=\{x:|f(x)|\le1\}\) and \(B=\{x:|f(x)|>1\}\). On \(A\), \(|f|^p\le1\), so \(\int_A|f|^p\,d\mu\le\mu(A)\le\mu(X)<\infty\). On \(B\), the strict inequality \(|f|>1\) and \(q>p\) imply \(|f|^p\le|f|^q\). Therefore \(\int_B|f|^p\,d\mu\le\int_B|f|^q\,d\mu\le\int_X|f|^q\,d\mu<\infty\). Adding the two integrals shows \(\int_X|f|^p\,d\mu<\infty\), which is exactly \(f\in L^p(\mu)\). If \(\mu(X)=0\), the same estimates hold and both integrals vanish. \(\square\)
The finite-total-measure hypothesis entered through the estimate on \(A\), where \(|f|^p\le1\). When \(\mu(X)=+\infty\), this bound alone no longer gives a finite integral. The theorem therefore makes no assertion in that setting. For constants and indicators, the earlier formulas give the more precise test: use \(\mu(X)\) for a nonzero constant and \(\mu(E)\) for \(\chi_E\).
A Common Mistake
A tempting but incorrect calculation is to say that the constant function \(f(x)=1\) on \(\mathbb{R}\) has \(L^2\) norm \(1\), because its value is \(1\). The norm is not just the value of the function: it is the square root of the integral of its square. For example, \(\int_{[-N,N]}|1|^2\,dm=2N\) for every positive integer \(N\), so the integral over \(\mathbb{R}\) is infinite and \(1\notin L^2(m)\). By contrast, on \([0,4]\) the same constant has squared integral \(1\cdot4=4\) and \(L^2\) norm \(2\).
A related error is to report \(\mu(E)\) as the \(L^p\) norm of an indicator. It is the integral of \(|\chi_E|^p\); the norm is its \(p\)th root, \(\mu(E)^{1/p}\), when \(\mu(E)<\infty\). For instance, if \(\mu(E)=4\) and \(p=2\), the integral is \(4\), but the norm is \(\sqrt{4}=2\).
The practical rule is to check the relevant measure before computing a norm. For a nonzero constant, that measure is the measure of the whole space. For an indicator, it is the measure of the indicated set. Infinite total measure does not prevent indicators of finite-measure sets from belonging to \(L^p\), and finite total measure does not make every measurable function integrable to every power.
Check Your Understanding
For each question, specify the measure space and exponent, and distinguish a finite norm from an infinite integral.
- On a space with \(\mu(X)=5\), compute the \(L^3(\mu)\) norm of the constant function \(f(x)=2\).
- On \(\mathbb{R}\) with Lebesgue measure, is \(\chi_{[2,7)}\) in \(L^p(m)\) for finite \(p\ge1\)? Compute its norm.
- Under counting measure on \(\mathbb{N}\), determine whether the indicator of the multiples of \(4\) belongs to \(L^1\).
- Explain why the zero constant belongs to \(L^p(\mu)\) when \(\mu(X)=+\infty\).
- For finite \(\mu(X)\) and \(1\le p<q<\infty\), identify the estimate on \(\{|f|\le1\}\) that is used to prove \(L^q(\mu)\subseteq L^p(\mu)\).