The Representation to Be Proved
For a measure space \((X,\mathcal{M},\mu)\), \(L^1(\mu)\) consists of equivalence classes, modulo equality \(\mu\)-a.e., of measurable real- or complex-valued functions \(f\) with \(\int_X |f|\,d\mu<\infty\). A bounded linear functional \(\Lambda\) on \(L^1(\mu)\) is a linear map to the same scalar field for which \(|\Lambda(f)|\le C\|f\|_1\) for some finite \(C\) and every \(f\in L^1(\mu)\). Its norm is \(\|\Lambda\|=\sup_{\|f\|_1\le1}|\Lambda(f)|\). No finiteness or completeness of the measure space is assumed below; sigma-finiteness is the needed measure-space hypothesis.
Here \(\|g\|_\infty\) is the essential supremum of \(|g|\): it is the least \(C\ge0\) such that \(|g|\le C\) \(\mu\)-a.e. The proof builds a measure from \(\Lambda\), applies the Radon–Nikodym theorem on finite-measure pieces, and then checks that the local densities fit together.
From a Functional to Local Measures
Choose measurable sets \(X_n\) with \(X_n\subseteq X_{n+1}\), \(\mu(X_n)<\infty\), and \(\bigcup_{n=1}^\infty X_n=X\); such an exhaustion exists by sigma-finiteness. Write \(C=\|\Lambda\|\). For each \(n\), define a set function on \(\mathcal{M}\) by $$ \nu_n(A)=\Lambda(\chi_{A\cap X_n}). $$ The indicator is in \(L^1(\mu)\), since \(\mu(A\cap X_n)\le\mu(X_n)<\infty\). In particular, $$ |\nu_n(A)|\le C\,\mu(A\cap X_n). $$
The set function \(\nu_n\) is countably additive. Indeed, if \(A_j\) are pairwise disjoint measurable sets and \(A=\bigcup_j A_j\), then the partial sums \(\sum_{j=1}^N\chi_{A_j\cap X_n}\) converge in \(L^1(\mu)\) to \(\chi_{A\cap X_n}\): the norm of the difference is \(\mu((\bigcup_{j>N}A_j)\cap X_n)\), which tends to zero by continuity from above inside the finite-measure set \(X_n\). Continuity of \(\Lambda\) therefore gives \(\nu_n(A)=\sum_j\nu_n(A_j)\). The bound above also shows that \(\nu_n\) has finite total variation and is absolutely continuous with respect to \(\mu\).
Apply the Radon–Nikodym theorem for finite signed or complex measures to \(\nu_n\) on \((X,\mathcal{M})\). There is an integrable measurable function \(g_n\), zero outside \(X_n\) up to a null set, such that \(\nu_n(A)=\int_A g_n\,d\mu\) for every \(A\in\mathcal{M}\). This use of Radon–Nikodym is the standard result established in the earlier material; the measure \(\nu_n\) is finite because \(|\nu_n|(X)\le C\mu(X_n)\).
We next verify the crucial bound \(|g_n|\le C\) a.e. On simple functions \(h\) supported in \(X_n\), linearity and the definition of \(\nu_n\) give \(\Lambda(h)=\int h g_n\,d\mu\). This identity also holds for every bounded measurable \(h\) supported in \(X_n\). To see this, choose simple functions \(h_j\), supported in \(X_n\), with \(\sup_j\|h_j\|_\infty\le\|h\|_\infty+1\) and \(h_j\to h\) pointwise; finite-valued quantization of the real and imaginary parts gives such approximations. Since \(\mu(X_n)<\infty\), bounded convergence gives \(h_j\to h\) in \(L^1\), so \(\Lambda(h_j)\to\Lambda(h)\). Also, dominated convergence with dominating function \((\|h\|_\infty+1)|g_n|\) gives \(\int h_jg_n\,d\mu\to\int hg_n\,d\mu\). Thus the identity passes to \(h\).
For \(\varepsilon>0\), let \(B=\{x\in X_n:|g_n(x)|>C+\varepsilon\}\). If \(\mu(B)>0\), use the bounded measurable function \(h=\chi_B\overline{g_n}/|g_n|\) on \(B\), defined to be zero elsewhere (in the real case, use the sign of \(g_n\)). The identity just proved and the functional bound imply $$ \int_B |g_n|\,d\mu=|\Lambda(h)|\le C\|h\|_1=C\mu(B). $$ But \(\int_B|g_n|\,d\mu\ge(C+\varepsilon)\mu(B)>C\mu(B)\), a contradiction. Hence \(\mu(B)=0\) for every \(\varepsilon>0\); taking \(\varepsilon=1/k\), \(k\ge1\), proves \(|g_n|\le C\) a.e.
Making the Densities Consistent
If \(m\ge n\), then for every measurable \(A\subseteq X_n\) we have \(\nu_m(A)=\Lambda(\chi_A)=\nu_n(A)\). The two Radon–Nikodym densities therefore agree a.e. on \(X_n\), by uniqueness in the Radon–Nikodym theorem. Put \(D_1=X_1\) and \(D_n=X_n\setminus X_{n-1}\) for \(n\ge2\). Define \(g=g_n\) on \(D_n\), assigning any value, say zero, on any exceptional null sets needed to choose measurable representatives. The \(D_n\) are disjoint measurable sets whose union is \(X\). The agreement on overlaps means that, for every fixed \(n\), \(g=g_n\) a.e. on \(X_n\). Since there are only countably many pieces, \(|g|\le C\) a.e. on \(X\), so \(g\in L^\infty(\mu)\).
Proof. First take a simple function \(s\) supported in \(X_n\). Its defining finite linear combination of indicators gives \(\Lambda(s)=\int s g_n\,d\mu=\int s g\,d\mu\). Next take any integrable simple function \(s\), not necessarily supported in one \(X_n\). Since \(|s|\) is integrable, \(s\chi_{X_n}\to s\) in \(L^1\), by dominated convergence applied to \(|s|\chi_{X\setminus X_n}\), which tends pointwise to zero and is bounded by \(|s|\). The identity holds for \(s\chi_{X_n}\); continuity of \(\Lambda\) and the bound \(|g|\le C\) give $$ \Lambda(s\chi_{X_n})\longrightarrow\Lambda(s),\qquad \left|\int_X(s\chi_{X_n}-s)g\,d\mu\right|\le C\|s\chi_{X_n}-s\|_1\longrightarrow0. $$ Thus the identity holds for every integrable simple \(s\). Finally, integrable simple functions are dense in \(L^1(\mu)\). For \(f\in L^1(\mu)\), choose such \(s_j\) with \(\|s_j-f\|_1\to0\). Then \(\Lambda(s_j)\to\Lambda(f)\), and $$ \left|\int_X(s_j-f)g\,d\mu\right|\le C\|s_j-f\|_1\longrightarrow0. $$ Passing to the limit proves the claim. \(\square\)
Uniqueness and the Exact Norm
Suppose \(g,h\in L^\infty(\mu)\) both represent \(\Lambda\). Let \(q=g-h\). Then \(\int f q\,d\mu=0\) for every \(f\in L^1(\mu)\). On \(X_n\), choose \(f=\chi_{X_n}\overline q/|q|\) where \(q\ne0\), and zero where \(q=0\). This is integrable because \(\mu(X_n)<\infty\), and the representation gives \(\int_{X_n}|q|\,d\mu=0\). Therefore \(q=0\) a.e. on each \(X_n\), and hence on \(X\). This proves uniqueness.
For the upper norm bound, the representation and \(|g|\le\|g\|_\infty\) a.e. imply $$ |\Lambda(f)|\le\int_X|f||g|\,d\mu\le\|g\|_\infty\|f\|_1, $$ so \(\|\Lambda\|\le\|g\|_\infty\). For the reverse bound, let \(0\le a<\|g\|_\infty\). By the definition of essential supremum, \(E=\{|g|>a\}\) has positive measure. Since \(X=\bigcup_nX_n\), some \(E\cap X_n\) has positive measure; it has finite measure as well. Set \(f=\chi_{E\cap X_n}\overline g/|g|\), with value zero where \(g=0\). Then \(\|f\|_1=\mu(E\cap X_n)\) and $$ |\Lambda(f)|=\int_{E\cap X_n}|g|\,d\mu\ge a\,\mu(E\cap X_n). $$ Thus \(\|\Lambda\|\ge a\). Letting \(a\) increase to \(\|g\|_\infty\) proves \(\|\Lambda\|\ge\|g\|_\infty\); if \(\|g\|_\infty=0\), the upper bound already gives equality. The theorem follows. \(\square\)
Worked Examples and a Common Pitfall
Worked Example: A Functional on a Three-Point Space
Let \(X=\{1,2,3\}\) with counting measure, and define \(\Lambda(f)=2f(1)-f(2)+\tfrac12 f(3)\). The representing function is \(g(1)=2\), \(g(2)=-1\), \(g(3)=\tfrac12\), because $$ \int_X fg\,d\mu=2f(1)-f(2)+\tfrac12 f(3)=\Lambda(f). $$ The essential-supremum norm is \(\|g\|_\infty=\max\{2,1,\tfrac12\}=2\). For any \(f\), the triangle inequality gives \(|\Lambda(f)|\le2(|f(1)|+|f(2)|+|f(3)|)=2\|f\|_1\). Taking \(f(1)=1\) and \(f(2)=f(3)=0\) gives \(\|f\|_1=1\) and \(|\Lambda(f)|=2\), so \(\|\Lambda\|=2\), as asserted.
Worked Example: A Density Whose Norm Is Not Attained by One Test Function
On \([0,1]\) with Lebesgue measure, let \(\Lambda(f)=\int_0^1 f(x)(1+x)\,dx\). The density \(g(x)=1+x\) has essential supremum \(2\), so the theorem gives \(\|\Lambda\|=2\). To see the norm approached by explicit unit vectors, take \(f_n=n\chi_{(1-1/n,1)}\), for \(n\ge2\). Its \(L^1\) norm is \(n(1/n)=1\), and direct integration gives $$ \Lambda(f_n)=n\int_{1-1/n}^1(1+x)\,dx =n\left[\left(x+\frac{x^2}{2}\right)\right]_{1-1/n}^1 =2-\frac{1}{2n}. $$ For \(n=2\) this value is \(2-\tfrac14=\tfrac74\), and as \(n\to\infty\) it tends to \(2\). Since every value is at most the norm \(2\), this independently verifies the norm computation through a sequence of unit-norm tests.
Worked Example: A Bounded Functional on an Infinite Counting Space
On \(\mathbb{N}=\{1,2,\ldots\}\) with counting measure, define \(\Lambda(a)=\sum_{k=1}^\infty (-1)^k a_k\) for \(a\in\ell^1\). The series converges absolutely because \(\sum_k|(-1)^k a_k|=\sum_k|a_k|<\infty\). The density \(g_k=(-1)^k\) belongs to \(\ell^\infty\) with norm \(1\), and \(|\Lambda(a)|\le\sum_k|a_k|=\|a\|_1\). For the sequence \(a_1=1\), \(a_k=0\) for \(k\ge2\), we have \(\|a\|_1=1\) and \(|\Lambda(a)|=1\), so the functional norm is exactly \(1\). As a further check, for \(a_k=2^{-k}\), the value is \(\sum_{k=1}^\infty(-1)^k2^{-k}=-\frac{1/2}{1+1/2}=-\tfrac13\), agreeing with the geometric series formula.
Common pitfall: L1 convergence does not by itself justify convergence of pairings. On \((0,1)\) with Lebesgue measure, let \(g(x)=x^{-1/2}\), which is integrable since \(\int_0^1x^{-1/2}\,dx=2\), and let \(u_n(x)=\sqrt n\,\chi_{(0,1/n)}(x)\). Then \(\|u_n\|_1=\sqrt n(1/n)=1/\sqrt n\to0\), but $$ \int_0^1u_n(x)g(x)\,dx =\sqrt n\int_0^{1/n}x^{-1/2}\,dx =\sqrt n\cdot\frac{2}{\sqrt n}=2. $$ Thus convergence in \(L^1\) alone does not ensure convergence of integrals against an arbitrary \(L^1\) function. In the local-density argument, the bounded measurable function \(u\) is instead approximated by simple functions that are uniformly bounded and converge pointwise; dominated convergence with the integrable majorant \((\|u\|_\infty+1)|g_n|\) justifies passage to the limit.
A related exhaustion error is to assume that every finite-measure set lies inside some \(X_n\). This need not hold: for \(X=[0,\infty)\), \(X_n=[0,n]\), the set \(A=\bigcup_{k\ge1}[k,k+2^{-k}]\) has finite measure \(\sum_{k\ge1}2^{-k}=1\), but is not contained in any \(X_n\). The proof avoids this error: for an integrable simple \(s\), it uses \(s\chi_{X_n}\), whose support is in \(X_n\), and then passes to the limit in \(L^1\).
Check Your Understanding
Use the representation theorem and its hypotheses to answer these questions.
- Why does \(\chi_{A\cap X_n}\) belong to \(L^1(\mu)\) even when \(\mu(A)\) is infinite?
- In the proof that \(|g_n|\le C\), what test function detects the phase of a complex-valued density?
- Why is agreement of the local densities only required a.e. on each \(X_n\), and why does countability of the exhaustion matter?
- For counting measure on \(\mathbb{N}\), what is the representing density of the functional \(a\mapsto\sum_{k\ge1}c_ka_k\) when \((c_k)\) is bounded?
- Which convergence argument justifies passing from \(s\chi_{X_n}\) to an integrable simple function \(s\) that is not supported in any one \(X_n\)?