From Finite Measure to Sigma-Finite Measure
On a finite-measure space, a bounded linear functional on \(L^1\) can be represented by integration against an essentially bounded function. A sigma-finite space may have infinite total measure, so that result cannot be applied to the whole space at once. Instead, we apply it on countably many finite-measure pieces, show that the local representations fit together, and pass from functions supported on finitely many pieces to arbitrary \(L^1\) functions.
Throughout, \((X,\mathcal{M},\mu)\) is a sigma-finite measure space: there are measurable sets of finite measure whose countable union is \(X\). Functions are complex-valued and measurable, and \(L^1(\mu)\) denotes the space of equivalence classes, modulo equality \(\mu\)-a.e., of functions \(f\) with \(\int_X |f|\,d\mu<\infty\). Its norm is \(\|f\|_1=\int_X|f|\,d\mu\). The function space \(L^\infty(\mu)\) consists of measurable functions that are bounded outside a \(\mu\)-null set, again considered up to equality \(\mu\)-a.e.; its norm is the essential supremum \(\|h\|_\infty\). A bounded linear functional \(\Lambda:L^1(\mu)\to\mathbb{C}\) is complex-linear and has norm \(\|\Lambda\|=\sup_{\|f\|_1\le1}|\Lambda(f)|<\infty\).
The Sigma-Finite Representation Theorem
Proof. Choose measurable sets \(A_j\) of finite measure with \(\bigcup_{j=1}^\infty A_j=X\). Make them disjoint by setting \(E_1=A_1\) and \(E_j=A_j\setminus\bigcup_{i=1}^{j-1}A_i\) for \(j\ge2\). Then the \(E_j\) are measurable, pairwise disjoint, have finite measure, and have union \(X\). Some pieces may be empty or null; no positive-measure assumption on a piece is needed.
For each \(j\), restrict \(\Lambda\) to functions supported on \(E_j\): for \(u\in L^1(E_j)\), extend \(u\) by zero to \(X\) and define \(\Lambda_j(u)=\Lambda(u)\). The extension has the same \(L^1\) norm, so \(|\Lambda_j(u)|\le\|\Lambda\|\|u\|_1\). Since \(\mu(E_j)<\infty\), the finite-measure representation result gives \(h_j\in L^\infty(E_j)\) with $$ \Lambda_j(u)=\int_{E_j}u h_j\,d\mu,\qquad \|h_j\|_{\infty,E_j}=\|\Lambda_j\|\le\|\Lambda\|. $$ Choose measurable representatives of the \(h_j\), and define \(h(x)=h_j(x)\) on \(E_j\). This piecewise definition is measurable because the \(E_j\) are a measurable partition. For each \(j\), the set in \(E_j\) where \(|h_j|>\|\Lambda\|\) is null; the union of these exceptional sets is a countable union of null sets and is null. Thus \(h\in L^\infty(\mu)\) and \(\|h\|_\infty\le\|\Lambda\|\).
Let \(F_N=\bigcup_{j=1}^N E_j\). For any \(f\in L^1(\mu)\), the function \(f\chi_{F_N}\) is supported on finitely many pieces. Linearity and the local representations give $$ \Lambda(f\chi_{F_N})=\sum_{j=1}^N\int_{E_j}f h\,d\mu=\int_{F_N}f h\,d\mu. $$ Also \(\|f-f\chi_{F_N}\|_1=\int_{X\setminus F_N}|f|\,d\mu\to0\). Indeed, the sets \(X\setminus F_N\) decrease to the empty set, and the integrable function \(|f|\) dominates \(|f|\chi_{X\setminus F_N}\), which tends pointwise to zero. The Dominated Convergence Theorem therefore gives the asserted limit. Boundedness of \(\Lambda\) implies \(\Lambda(f\chi_{F_N})\to\Lambda(f)\). Since \(h\) is essentially bounded, $$ \left|\int_X f h\,d\mu-\int_{F_N}f h\,d\mu\right| \le \|h\|_\infty\int_{X\setminus F_N}|f|\,d\mu\longrightarrow0. $$ Taking limits in the finite-piece identity proves \(\Lambda(f)=\int_X f h\,d\mu\) for every \(f\in L^1(\mu)\).
For uniqueness, suppose \(h,k\in L^\infty(\mu)\) both represent \(\Lambda\), and set \(d=h-k\). Then \(\int f d\,d\mu=0\) for every \(f\in L^1(\mu)\). On a piece \(E_j\), define \(f=\chi_{E_j}\overline d/|d|\) where \(d\ne0\), and \(f=0\) where \(d=0\). This is measurable and belongs to \(L^1\), since \(|f|\le\chi_{E_j}\) and \(\mu(E_j)<\infty\). Hence $$ 0=\int_X f d\,d\mu=\int_{E_j}|d|\,d\mu. $$ A nonnegative measurable function with integral zero is zero almost everywhere, so \(d=0\) a.e. on \(E_j\). The countable union of the \(E_j\) is \(X\), and therefore \(h=k\) \(\mu\)-a.e.
Finally, \(|\Lambda(f)|\le\|h\|_\infty\|f\|_1\), so \(\|\Lambda\|\le\|h\|_\infty\). If \(\|h\|_\infty=0\), this already gives equality. Otherwise, for any \(a\) with \(0\le a<\|h\|_\infty\), the set \(\{|h|>a\}\) has positive measure. Since \(X=\bigcup_jE_j\), some \(E=E_j\cap\{|h|>a\}\) has positive measure. It has finite measure because \(\mu(E_j)<\infty\). Set \(f=\chi_E\overline h/|h|\,/\,\mu(E)\) on \(E\), and zero outside \(E\). Then \(\|f\|_1=1\) and $$ |\Lambda(f)|=\frac{1}{\mu(E)}\int_E|h|\,d\mu>a. $$ Thus \(\|\Lambda\|\ge a\) for every \(a<\|h\|_\infty\), which yields \(\|\Lambda\|\ge\|h\|_\infty\). Combined with the upper bound, this proves the norm identity. \(\square\)
How the Local Representations Fit Together
Worked Example: A Functional on Counting Measure
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), let \(\mathcal{M}\) be all subsets of \(\mathbb{N}\), and let \(\mu\) be counting measure. This space is sigma-finite because each singleton has measure \(1\). Define \(h(k)=1\) when \(k\) is even and \(h(k)=-\tfrac12\) when \(k\) is odd, and set \(\Lambda(x)=\sum_{k=1}^\infty x_k h(k)\) for \(x\in L^1(\mu)=\ell^1\).
For every \(x\in\ell^1\), the series is absolutely convergent because $$ \sum_{k=1}^\infty|x_kh(k)| \le\sum_{k=1}^\infty|x_k|=\|x\|_1. $$ Thus \(\Lambda\) is well-defined and linear, and \(|\Lambda(x)|\le\|x\|_1\), so \(\|\Lambda\|\le1\). For the sequence \(e_2\), which is \(1\) at \(k=2\) and \(0\) elsewhere, \(\|e_2\|_1=1\) and \(\Lambda(e_2)=1\). Hence \(\|\Lambda\|\ge1\), so \(\|\Lambda\|=1\). The representing function is \(h\), with \(\|h\|_\infty=1\): \(|h(k)|\le1\) at every \(k\), and \(|h(2)|=1\). On the singleton partition \(E_k=\{k\}\), the local representation is multiplication by \(h(k)\); gluing these values gives precisely the displayed \(h\). The norm calculation agrees with the theorem.
Worked Example: Combining Pieces on the Real Line
Let \((\mathbb{R},\mathcal{L},m)\) be the Lebesgue measure space, which is sigma-finite. Use the partition \([j,j+1)\), \(j\in\mathbb{Z}\), into sets of measure \(1\). Define \(h=2\) on \([0,1)\), \(h=-1\) on \([2,3)\), and \(h=0\) elsewhere. The functional \(\Lambda(f)=\int_{\mathbb{R}} f h\,dm\) is bounded on complex-valued \(L^1(m)\), since \(|\Lambda(f)|\le2\|f\|_1\).
Take \(f=2\chi_{[0,1)}-\chi_{[2,4)}\). Its absolute value is \(2\) on \([0,1)\), \(1\) on \([2,4)\), and zero elsewhere. Therefore \(\|f\|_1=2\cdot1+1\cdot2=4\). The product \(fh\) is \(4\) on \([0,1)\), \(1\) on \([2,3)\), and \(0\) elsewhere, so $$ \Lambda(f)=\int_{\mathbb{R}}fh\,dm=4\cdot1+1\cdot1=5. $$ This also follows by splitting into partition pieces: the contribution from \([0,1)\) is \(2\cdot2\cdot1=4\), from \([2,3)\) it is \((-1)(-1)\cdot1=1\), and from \([3,4)\) it is \((-1)\cdot0\cdot1=0\). The total is \(4+1+0=5\), as computed directly. The function \(h\) has essential supremum \(2\), since \(|h|\le2\) everywhere and \(|h|=2\) on a set of measure \(1\).
Worked Example: Passing from Truncations to the Whole Space
On \((\mathbb{R},\mathcal{L},m)\), let \(h(x)=1\) and \(f(x)=e^{-|x|}\). Then \(f\in L^1(m)\), and \(\Lambda(f)=\int_{\mathbb{R}}f h\,dm\) is represented by the bounded function \(h\). For \(F_N=[-N,N)\), with \(N\) a positive integer, $$ \int_{F_N}f\,dm=2\int_0^N e^{-x}\,dx=2(1-e^{-N}). $$ The \(L^1\) tail is $$ \int_{\mathbb{R}\setminus F_N}f\,dm=2\int_N^\infty e^{-x}\,dx=2e^{-N}\longrightarrow0. $$ Consequently \(\int_{\mathbb{R}} f\,dm=\lim_{N\to\infty}2(1-e^{-N})=2\), and the truncation identity passes to the full integral. As a check, the two half-line integrals are each \(\int_0^\infty e^{-x}\,dx=1\), giving \(1+1=2\).
A Common Pitfall: Adding the Local Norms
The local representations are glued by defining one function on each piece, not by summing the norms of the local functionals. For example, on \(\mathbb{N}\) with counting measure, take \(h(k)=1\) for every \(k\) and \(\Lambda(x)=\sum_{k=1}^\infty x_k\). The restriction to each singleton has norm \(1\), so the sum of the local norms is \(\sum_{k=1}^\infty1=\infty\). But globally, $$ |\Lambda(x)|\le\sum_{k=1}^\infty|x_k|=\|x\|_1, $$ and for \(e_1\) we have \(\|e_1\|_1=1\) and \(\Lambda(e_1)=1\). Thus \(\|\Lambda\|=1\), not infinity; its representing function is the constant function \(1\), with \(\|h\|_\infty=1\). The local norm bound is uniform across pieces, and the essential supremum of the glued density—not the sum of local norms—determines the global norm.
Check Your Understanding
Use the partition argument and the stated finite-measure representation result to organize your reasoning.
- Starting from a sigma-finite cover \(A_j\), write the disjointification that produces a measurable partition into finite-measure sets.
- Why does the finite-measure representation theorem apply to the restriction of \(\Lambda\) to functions supported on a single partition piece?
- Explain why the local bounds \(\|h_j\|_\infty\le\|\Lambda\|\) imply that the glued function belongs to \(L^\infty(\mu)\).
- Which convergence theorem justifies \(\|f-f\chi_{F_N}\|_1\to0\), and what are its dominating function and pointwise limit?
- Why is it valid to test uniqueness using a function supported on one finite-measure piece, even when \(\mu(X)=\infty\)?