Representing L1 Functionals on Finite Measure Spaces

From a Functional to a Measure

Let \((X,\mathcal{M},\mu)\) be a finite measure space, meaning \(\mu(X)<\infty\). We work with complex-valued measurable functions and identify functions equal \(\mu\)-almost everywhere. The space \(L^1(\mu)\) consists of those equivalence classes \(f\) for which \(\|f\|_1=\int_X|f|\,d\mu<\infty\). A bounded complex-linear functional \(\Lambda:L^1(\mu)\to\mathbb{C}\) has operator norm \(\|\Lambda\|=\sup_{\|f\|_1\le1}|\Lambda(f)|\).

Since \(\mu\) is finite, every indicator \(\chi_E\), \(E\in\mathcal{M}\), belongs to \(L^1(\mu)\). Define \(\nu(E)=\Lambda(\chi_E)\). The crucial point is that this set function is a finite complex measure, so its density can be obtained from the complex Radon–Nikodym theorem.

Proof. If \(E_1,E_2,\ldots\) are pairwise disjoint measurable sets and \(E=\bigcup_{j\ge1}E_j\), then \(\chi_{\cup_{j=1}^nE_j}\to\chi_E\) in \(L^1(\mu)\): the norm of the difference is \(\mu(\bigcup_{j>n}E_j)\), which tends to zero by countable additivity and \(\mu(E)\le\mu(X)<\infty\). Boundedness of \(\Lambda\) gives \(\nu(E)=\lim_n\sum_{j=1}^n\nu(E_j)=\sum_{j\ge1}\nu(E_j)\). For any finite measurable partition \(E_1,\ldots,E_r\) of \(E\), choose complex numbers \(c_j\) with \(|c_j|=1\) and \(c_j\nu(E_j)=|\nu(E_j)|\) whenever \(\nu(E_j)\ne0\); when it is zero, choose \(c_j=1\). Then \[ \sum_{j=1}^r|\nu(E_j)| =\left|\Lambda\left(\sum_{j=1}^r c_j\chi_{E_j}\right)\right| \le \|\Lambda\|\sum_{j=1}^r\mu(E_j) =\|\Lambda\|\mu(E). \] Thus the total variation satisfies \(|\nu|(E)\le\|\Lambda\|\mu(E)\), in particular \(|\nu|(X)<\infty\). If \(\mu(E)=0\), then \(\|\chi_E\|_1=0\), so \(\nu(E)=0\). \(\square\)

The Representation Theorem

The complex Radon–Nikodym theorem states that if a complex measure \(\nu\) has finite total variation and is absolutely continuous with respect to a sigma-finite positive measure \(\mu\), then there is an integrable complex-valued measurable function \(h\) such that \(\nu(E)=\int_Eh\,d\mu\) for every measurable \(E\). The sigma-finiteness hypothesis is essential to the theorem as stated. Here it holds because the finite measure \(\mu\) is sigma-finite: \(X=\bigcup_{n\ge1}X\) and \(\mu(X)<\infty\). We use this standard theorem, rather than prove the complex Radon–Nikodym theorem here.

Proof. Form \(\nu(E)=\Lambda(\chi_E)\) as above. It is a complex measure of finite total variation, absolutely continuous with respect to the finite, hence sigma-finite, measure \(\mu\). The complex Radon–Nikodym theorem supplies an integrable complex-valued \(h\) with \(\nu(E)=\int_Eh\,d\mu\).

Write \(C=\|\Lambda\|\). We show \(|h|\le C\) almost everywhere. For any \(E\in\mathcal{M}\), define \(f_E=\chi_E\overline{h}/|h|\) where \(h\ne0\), and set \(f_E=0\) elsewhere. This is measurable, and \(\|f_E\|_1=\mu(E\cap\{|h|>0\})\le\mu(E)\). The representation is not yet known for \(f_E\), so instead use the finite-measure bound on \(\nu\) proved above: \(|\nu|(E)\le C\mu(E)\). The variation of a measure with density \(h\) is \(|\nu|(E)=\int_E|h|\,d\mu\), the standard variation formula for complex measures with an integrable density. Consequently \(\int_E|h|\,d\mu\le C\mu(E)\) for every measurable \(E\). If the set \(A=\{|h|>C\}\) had positive measure, then some \(A_n=\{|h|\ge C+1/n\}\) would have positive measure, since \(A=\bigcup_{n\ge1}A_n\). For that set, \[ \int_{A_n}|h|\,d\mu\ge(C+1/n)\mu(A_n)>C\mu(A_n), \] a contradiction. Hence \(|h|\le C\) a.e., so \(h\in L^\infty(\mu)\).

For a complex simple function \(s=\sum_{j=1}^r a_j\chi_{E_j}\) with disjoint measurable \(E_j\), linearity and the definition of \(\nu\) give \[ \Lambda(s)=\sum_{j=1}^r a_j\nu(E_j)=\sum_{j=1}^r a_j\int_{E_j}h\,d\mu=\int_X sh\,d\mu. \] Complex integrable simple functions are dense in \(L^1(\mu)\), a standard approximation result for the Lebesgue integral. Given \(f\in L^1(\mu)\), choose simple \(s_n\) with \(\|s_n-f\|_1\to0\). Boundedness of \(\Lambda\) implies \(\Lambda(s_n)\to\Lambda(f)\); since \(|h|\le C\) a.e., \[ \left|\int_X(s_n-f)h\,d\mu\right|\le C\|s_n-f\|_1\longrightarrow0. \] Therefore \(\Lambda(f)=\int_Xfh\,d\mu\) for every \(f\in L^1(\mu)\). This also gives \(|\Lambda(f)|\le\|h\|_\infty\|f\|_1\), hence \(\|\Lambda\|\le\|h\|_\infty\).

For the reverse norm inequality, let \(0\le a<\|h\|_\infty\). By the definition of essential supremum, \(E=\{|h|>a\}\) has positive measure. Since \(\mu\) is finite, \(f=\chi_E\overline{h}/|h|\), defined as zero where \(h=0\), lies in \(L^1(\mu)\) and \(\|f\|_1=\mu(E)>0\). The representation just proved yields \(\Lambda(f)=\int_E|h|\,d\mu\ge a\mu(E)\). Thus \(\|\Lambda\|\ge|\Lambda(f)|/\|f\|_1\ge a\). Letting \(a\) increase to \(\|h\|_\infty\) proves \(\|\Lambda\|\ge\|h\|_\infty\), including the case \(\|h\|_\infty=0\), where the reverse inequality is automatic.

Finally, if \(h\) and \(k\) both represent \(\Lambda\), then \(\int_E(h-k)\,d\mu=0\) for every measurable \(E\), by testing \(f=\chi_E\). Put \(q=h-k\). For each positive integer \(n\), the real-valued function \(\operatorname{Re}q\) cannot be at least \(1/n\) on a positive-measure set \(B\), because then \(\operatorname{Re}\int_Bq\,d\mu=\int_B\operatorname{Re}q\,d\mu\ge\mu(B)/n>0\). Applying this to \(\{\operatorname{Re}q\ge1/n\}\) and \(\{\operatorname{Re}q\le-1/n\}\) shows \(\operatorname{Re}q=0\) a.e. The same argument using the imaginary part shows \(\operatorname{Im}q=0\) a.e. Hence \(h=k\) a.e. \(\square\)

What the Norm Measures—and a Common Pitfall

The operator norm is the essential supremum of the density, not the total variation of the induced measure. The distinction is visible on \(([0,1],\mathcal{L},m)\) for \(h(x)=x\) and \(\Lambda(f)=\int_0^1fx\,dm\). The density has \(\|h\|_\infty=1\), so \(\|\Lambda\|=1\). In contrast, the total variation of the induced measure is \(|\nu|([0,1])=\int_0^1x\,dm=1/2\). To check the operator norm directly, for \(0<\varepsilon<1\), set \(E_\varepsilon=(1-\varepsilon,1]\) and \(f_\varepsilon=\chi_{E_\varepsilon}/\varepsilon\). Then \(\|f_\varepsilon\|_1=1\) and \[ \Lambda(f_\varepsilon)=\frac1\varepsilon\int_{1-\varepsilon}^1x\,dx =\frac{1-(1-\varepsilon)^2}{2\varepsilon}=1-\frac{\varepsilon}{2}. \] These values tend to \(1\), confirming the norm. The integral of the density, \(1/2\), instead gives the total variation, not the functional norm.

A related pitfall is to omit sigma-finiteness when invoking the complex Radon–Nikodym theorem. The finite-measure setting avoids that error: every finite measure is sigma-finite. It also ensures every indicator belongs to \(L^1\), which is needed to define the induced measure through \(\Lambda(\chi_E)\).