From a Functional to Values on Sets
In a finite measure space, every measurable set has an indicator function in \(L^1\). This makes it possible to ask what a bounded linear functional does to indicators and use the answers as the values of a set function. The important issue is countable additivity: it does not follow just from defining values on indicators, but from the continuity of the functional when a disjoint union is approximated by finite unions.
Let \((X,\mathcal{M},\mu)\) be a finite measure space, so \(\mu(X)<\infty\). No completeness assumption is needed. The space \(L^1(\mu)\) consists of equivalence classes, modulo equality \(\mu\)-a.e., of complex-valued measurable functions \(f\) for which \(\int_X |f|\,d\mu<\infty\), with norm \(\|f\|_1=\int_X|f|\,d\mu\). Suppose \(\Lambda:L^1(\mu)\to\mathbb{C}\) is complex-linear and bounded: there is a constant \(C<\infty\) such that \(|\Lambda(f)|\le C\|f\|_1\) for every \(f\in L^1(\mu)\). We may take \(C=\|\Lambda\|\), the operator norm.
The use of equivalence classes causes no ambiguity. If \(E,F\in\mathcal{M}\) and \(\mu(E\triangle F)=0\), then \(\|\chi_E-\chi_F\|_1=\mu(E\triangle F)=0\), so \(\Lambda(\chi_E)=\Lambda(\chi_F)\). In particular, every \(\mu\)-null set has \(\nu\)-value zero. This gives \(\nu\ll\mu\), meaning that \(\nu(E)=0\) whenever \(\mu(E)=0\). We will prove the stronger countable additivity and variation bounds below.
Countable Additivity from \(L^1\) Convergence
Proof. First, \(\nu(\varnothing)=\Lambda(0)=0\). Let \((E_n)_{n\ge1}\) be pairwise disjoint measurable sets and put \(A=\bigcup_{n\ge1}E_n\), \(A_N=\bigcup_{n=1}^N E_n\). Since \(\mu\) is a measure, countable additivity gives \(\mu(A)=\sum_{n\ge1}\mu(E_n)\le\mu(X)<\infty\). The indicators \(\chi_{A_N}\) converge to \(\chi_A\) in \(L^1(\mu)\), because $$ \|\chi_A-\chi_{A_N}\|_1 =\mu(A\setminus A_N) =\sum_{n>N}\mu(E_n)\longrightarrow 0. $$ The last limit holds because the nonnegative series has finite sum. Boundedness of \(\Lambda\) now gives $$ |\nu(A)-\Lambda(\chi_{A_N})| \le \|\Lambda\|\,\|\chi_A-\chi_{A_N}\|_1\longrightarrow 0. $$ By finite linearity and disjointness, \(\Lambda(\chi_{A_N})=\sum_{n=1}^N\Lambda(\chi_{E_n})=\sum_{n=1}^N\nu(E_n)\). Thus the partial sums converge to \(\nu(A)\), proving countable additivity.
It remains to prove the variation estimate. For a measurable set \(E\), total variation is given by the supremum of \(\sum_{j=1}^k|\nu(E_j)|\) over all finite measurable partitions \(E=\bigcup_{j=1}^k E_j\) into pairwise disjoint sets. Fix such a partition. For each \(j\) with \(\nu(E_j)\ne0\), put \(c_j=\overline{\nu(E_j)}/|\nu(E_j)|\); if \(\nu(E_j)=0\), put \(c_j=1\). Then \(|c_j|=1\) and \(c_j\nu(E_j)=|\nu(E_j)|\). The simple function \(g=\sum_{j=1}^k c_j\chi_{E_j}\) satisfies \(|g|=\chi_E\), so \(\|g\|_1=\mu(E)\). By complex-linearity and boundedness, $$ \sum_{j=1}^k|\nu(E_j)| =\Lambda(g) \le |\Lambda(g)| \le \|\Lambda\|\mu(E). $$ Here the sum on the left is a nonnegative real number, and \(\Lambda(g)=\sum_jc_j\nu(E_j)\) equals that number. Taking the supremum over the finite partitions proves \(|\nu|(E)\le\|\Lambda\|\mu(E)\). Setting \(E=X\) shows finite total variation, and the estimate also shows that \(\mu(E)=0\) implies \(|\nu|(E)=0\), hence \(\nu(E)=0\). \(\square\)
The finite measure assumption enters at the first definition: it ensures \(\chi_E\in L^1(\mu)\) for every measurable \(E\). Countable additivity then follows from a second finiteness fact, namely that the measures of the tails of a disjoint union tend to zero. The estimate on total variation is stronger than mere countable additivity: it controls all finite measurable partitions at once.
Worked Constructions
Worked Example: A Functional on a Three-Point Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be its power set, and assign masses \(\mu(\{a\})=1/2\), \(\mu(\{b\})=1/3\), and \(\mu(\{c\})=1/6\). Define \(\Lambda(f)=2i f(a)\mu(\{a\})-3f(b)\mu(\{b\})+(1+i)f(c)\mu(\{c\})\). This is bounded: each coefficient has modulus at most \(3\), so \(|\Lambda(f)|\le3\|f\|_1\).
Apply the definition to singleton indicators. For \(E=\{a\}\), \(\nu(E)=2i(1/2)=i\). For \(E=\{b\}\), \(\nu(E)=-3(1/3)=-1\). For \(E=\{c\}\), \(\nu(E)=(1+i)(1/6)=(1+i)/6\). Therefore, for \(E=\{a,c\}\), finite additivity gives $$ \nu(\{a,c\})=i+\frac{1+i}{6}=\frac{1+7i}{6}. $$ Directly, \(\chi_{\{a,c\}}\) is \(1\) at \(a,c\) and \(0\) at \(b\), so applying \(\Lambda\) gives \(2i(1/2)+(1+i)(1/6)=(1+7i)/6\), as required. The total variation on \(X\) is \(|i|+|-1|+|(1+i)/6|=2+\sqrt{2}/6\), which is at most the theorem's bound \(\|\Lambda\|\mu(X)\le3\cdot1=3\).
Worked Example: A Functional on the Unit Interval
Take \(X=[0,1]\), \(\mathcal{M}=\mathcal{L}|_{[0,1]}\), and \(\mu=m\), Lebesgue measure restricted to this interval. Define \(\Lambda(f)=\int_0^1 f(x)(1+2ix)\,dx\). Since \(|1+2ix|=\sqrt{1+4x^2}\le\sqrt5\) on \([0,1]\), we have \(|\Lambda(f)|\le\sqrt5\|f\|_1\). For \(E=[0,1/2]\), $$ \nu(E)=\Lambda(\chi_E) =\int_0^{1/2}(1+2ix)\,dx =\left[x+ix^2\right]_0^{1/2} =\frac12+\frac{i}{4}. $$ For \(F=(1/2,1]\), the same antiderivative gives \(\nu(F)=[x+ix^2]_{1/2}^1=\frac12+\frac{3i}{4}\). These sets are disjoint and their union is \([0,1]\); consequently \(\nu([0,1])=1+i\). Checking directly, \(\Lambda(\chi_{[0,1]})=\int_0^1(1+2ix)\,dx=[x+ix^2]_0^1=1+i\), agreeing with the sum of the two set values.
Worked Example: Countable Additivity on a Countable Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), \(\mathcal{M}\) its power set, and \(\mu(\{n\})=2^{-n}\). The geometric series gives \(\mu(X)=\sum_{n=1}^\infty2^{-n}=1\), so this is a finite measure space. Define \(\Lambda(f)=\sum_{n=1}^\infty (1+(-1)^n i)f(n)2^{-n}\). The series is absolutely convergent for \(f\in L^1(\mu)\), since \(|1+(-1)^ni|=\sqrt2\) and thus \(|\Lambda(f)|\le\sqrt2\sum_n|f(n)|2^{-n}=\sqrt2\|f\|_1\).
For each singleton, \(\nu(\{n\})=(1+(-1)^ni)2^{-n}\). The even integers have total mass \(\sum_{k=1}^\infty2^{-2k}=(1/4)/(1-1/4)=1/3\), so \(\nu(\{2,4,\ldots\})=(1+i)/3\). The odd integers have total mass \(\sum_{k=1}^\infty2^{-(2k-1)}=(1/2)/(1-1/4)=2/3\), so \(\nu(\{1,3,\ldots\})=2(1-i)/3\). Adding gives \(\nu(X)=1-i/3\). Direct substitution into \(\Lambda(\chi_X)\) yields the same value: the even terms sum to \((1+i)/3\), the odd terms to \(2(1-i)/3\), and their sum is \(1-i/3\). The calculation illustrates how the induced set values sum over a countable disjoint partition.
Common Pitfalls and the Role of the Hypotheses
Confusing finite additivity with countable additivity. Linearity immediately gives finite additivity for disjoint sets, but that alone does not establish a measure. For a countable disjoint union, the missing step is to show that the indicators of the finite unions converge in \(L^1\). The estimate \(\|\chi_A-\chi_{A_N}\|_1=\mu(A\setminus A_N)\to0\), followed by boundedness of \(\Lambda\), supplies exactly that step. Without continuity of the functional, finite additivity would not by itself justify passage from finite unions to their countable union.
Trying to use point evaluation on \(L^1\). On \([0,1]\) with Lebesgue measure, the rule \(\Lambda(f)=f(0)\) is not a well-defined functional on \(L^1\) equivalence classes. The function represented by \(f(x)=0\) for every \(x\) and the function represented by \(g=\chi_{\{0\}}\) are equal \(\mu\)-a.e., because \(\{0\}\) has measure zero. Yet the proposed rule assigns them values \(0\) and \(1\). Thus it does not define a map on \(L^1\), let alone a bounded one.
Forgetting why the space must be finite here. If \(\mu\) is Lebesgue measure on \(\mathbb{R}\), then \(\chi_{\mathbb{R}}\notin L^1(m)\), so \(\Lambda(\chi_E)\) is not automatically defined for every measurable set \(E\). On sets of finite measure, indicators are in \(L^1\), and the construction can be applied there; the present theorem makes a set function on all of \(\mathcal{M}\) because \(\mu(X)<\infty\). The result above constructs a measure from the functional; it does not yet identify a function that represents the functional on every element of \(L^1\).
Check Your Understanding
Use the definition on indicators and the \(L^1\) estimate to justify each answer.
- Why does \(\chi_E\) belong to \(L^1(\mu)\) for every \(E\in\mathcal{M}\) when \(\mu(X)<\infty\)?
- For disjoint \(E_n\), write the exact \(L^1\) norm that tends to zero when \(\chi_{\cup_{n=1}^N E_n}\) approaches \(\chi_{\cup_{n=1}^\infty E_n}\).
- If \(\|\Lambda\|\le4\) and \(\mu(E)=1/5\), what bound does the variation estimate give for \(|\nu|(E)\)?
- Explain why the proposed point-evaluation rule \(f\mapsto f(0)\) is not well-defined on \(L^1([0,1],m)\).
- For a disjoint partition \(E=\bigcup_{j=1}^kE_j\), how are the coefficients \(c_j\) chosen to turn \(\sum_j|\nu(E_j)|\) into the value of \(\Lambda\) on a simple function?