From Indicators to Bounded Functions
Many measure-theoretic identities are easiest to verify first for indicators: an indicator records whether a point belongs to a set, and an algebra of sets gives a manageable collection of such tests. A monotone class argument turns those tests into an identity for bounded measurable functions. The essential extra ingredient is continuity of the quantities in the identity under bounded increasing limits.
We work on a measurable space \((X,\mathcal{M})\). A function is called bounded if it is real-valued and there is a finite constant \(C\) such that \(|f(x)|\le C\) for every \(x\in X\). In the main result, all functions are bounded and \(\mathcal{M}\)-measurable, and all function identities hold everywhere. The functionals take finite real values.
For functions, the argument will use the following form of continuity: a functional \(T\) is continuous under bounded increasing limits if \(T(f_n)\to T(f)\) whenever \(0\le f_n\uparrow f\le C\) for a finite constant \(C\). If \(T(f)=\int f\,d\mu\) for a finite measure \(\mu\), this property follows from the Monotone Convergence Theorem. Finiteness ensures that the integrals of bounded functions are finite.
The Monotone Class Theorem for an Algebra
Proof. Let \(\mathcal{C}\) be the intersection of all monotone classes containing \(\mathcal{A}\). Intersections of monotone classes are monotone classes, so \(\mathcal{C}\) is the smallest one. Since \(\sigma(\mathcal{A})\) is a monotone class containing \(\mathcal{A}\), minimality gives \(\mathcal{C}\subseteq\sigma(\mathcal{A})\).
We show that \(\mathcal{C}\) is itself a sigma-algebra. Fix \(A\in\mathcal{A}\), and let \(\mathcal{D}_A=\{B\subseteq X:A\cap B\in\mathcal{C}\}\). This is a monotone class: increasing unions and decreasing intersections of sets \(B\) give the corresponding limits of \(A\cap B\), which belong to \(\mathcal{C}\). It contains \(\mathcal{A}\), because an algebra is closed under intersections. Thus \(\mathcal{C}\subseteq\mathcal{D}_A\). Consequently, \(A\cap B\in\mathcal{C}\) whenever \(A\in\mathcal{A}\) and \(B\in\mathcal{C}\).
Now fix \(B\in\mathcal{C}\), and consider \(\mathcal{E}_B=\{A\subseteq X:A\cap B\in\mathcal{C}\}\). The same limit argument shows that \(\mathcal{E}_B\) is a monotone class. The conclusion of the preceding paragraph says that \(\mathcal{A}\subseteq\mathcal{E}_B\), so \(\mathcal{C}\subseteq\mathcal{E}_B\). Therefore \(\mathcal{C}\) is closed under finite intersections. The class \(\{A\subseteq X:A^c\in\mathcal{C}\}\) is also a monotone class and contains \(\mathcal{A}\); hence it contains \(\mathcal{C}\). Thus \(\mathcal{C}\) is closed under complements as well as finite intersections. It contains \(X\), since \(X\in\mathcal{A}\), so it is an algebra. Closure under increasing unions now gives closure under countable unions: for any \(A_n\in\mathcal{C}\), the finite unions \(\bigcup_{k=1}^n A_k\) belong to \(\mathcal{C}\) and increase to \(\bigcup_{k=1}^\infty A_k\). Hence \(\mathcal{C}\) is a sigma-algebra containing \(\mathcal{A}\), so \(\sigma(\mathcal{A})\subseteq\mathcal{C}\). Combining the two inclusions proves the claim. \(\square\)
Extending an Identity of Functionals
Proof. Let \(\mathcal{C}\) be the collection of sets \(E\in\mathcal{M}\) for which \(L(\chi_E)=R(\chi_E)\). It contains \(\mathcal{A}\). If \(E_n\uparrow E\) with \(E_n\in\mathcal{C}\), then \(\chi_{E_n}\uparrow\chi_E\), with all functions bounded by \(1\). Continuity under bounded increasing limits gives \(L(\chi_E)=\lim_nL(\chi_{E_n})=\lim_nR(\chi_{E_n})=R(\chi_E)\). If \(E_n\downarrow E\), the indicators decrease rather than increase. In this case apply increasing-limit continuity to \(1-\chi_{E_n}\uparrow1-\chi_E\); linearity and \(L(1)=R(1)\), which holds because \(X\in\mathcal{A}\), give the same equality for \(E\). Thus \(\mathcal{C}\) is a monotone class. By the theorem just proved, it contains \(\sigma(\mathcal{A})\).
Linearity now gives \(L(s)=R(s)\) for every simple \(\sigma(\mathcal{A})\)-measurable function \(s\), since such a function is a finite linear combination of indicators of measurable sets. To pass to a bounded measurable \(f\), choose a constant \(c\) so that \(g=f+c\ge0\), and choose a finite \(C\) with \(g\le C\). The functions \(s_n=2^{-n}\lfloor 2^n g\rfloor\) are nonnegative simple \(\sigma(\mathcal{A})\)-measurable functions. They increase pointwise to \(g\), and \(0\le s_n\le C\). Hence \(L(g)=\lim_nL(s_n)=\lim_nR(s_n)=R(g)\). Subtracting \(cL(1)=cR(1)\) gives \(L(f)=R(f)\). \(\square\)
In this proof, positivity ensures the functionals behave appropriately on nonnegative functions; the extension itself uses linearity and continuity for increasing limits. For finite measures \(\mu\) and \(\nu\) on \((X,\mathcal{M})\), the functionals \(L(f)=\int f\,d\mu\) and \(R(f)=\int f\,d\nu\) meet these assumptions by the Monotone Convergence Theorem. The theorem is not restricted to integrals, however: it applies to any pair of positive linear functionals with the stated continuity.
Worked Applications
Worked Example: Reflection on the Unit Interval
Let \(m\) be Lebesgue measure on \([0,1]\), and let \(\mathcal{A}\) be the algebra of finite unions of intervals \([a,b)\subseteq[0,1]\), together with the singleton \(\{1\}\). It generates the Borel sigma-algebra on \([0,1]\). Show that for every bounded Borel-measurable real-valued \(f\),
Define \(L(f)=\int_0^1 f(x)\,dx\) and \(R(f)=\int_0^1 f(1-x)\,dx\). Both are positive linear functionals on bounded Borel-measurable functions and are continuous under bounded increasing limits, by the Monotone Convergence Theorem. For an interval \([a,b)\), its inverse image under \(x\mapsto1-x\) is \((1-b,1-a]\); its Lebesgue measure is \(b-a\), the same as the measure of \([a,b)\). The reflected image of \(\{1\}\) is \(\{0\}\), and both singletons have measure zero. Finite additivity therefore gives \(L(\chi_A)=R(\chi_A)\) for every \(A\in\mathcal{A}\). The theorem extends the identity to every bounded Borel-measurable \(f\).
Check: take \(f(x)=x\). Then \(L(f)=\int_0^1x\,dx=1/2\), while \(R(f)=\int_0^1(1-x)\,dx=1-1/2=1/2\).
Worked Example: Rescaling the Half-Interval
For every bounded Borel-measurable real-valued \(f\) on \([0,1]\), prove that
Use the same algebra \(\mathcal{A}\). Set \(L(f)=\int_0^1 f(y)\,dy\) and \(R(f)=2\int_0^{1/2}f(2x)\,dx\). These are positive linear functionals, and both are continuous under bounded increasing limits by the Monotone Convergence Theorem. For \(A=[a,b)\), where \(0\le a<b\le1\), the set of \(x\in[0,1/2]\) for which \(2x\in[a,b)\) is \([a/2,b/2)\), up to a possible endpoint of measure zero. Its measure is \((b-a)/2\), so \(R(\chi_A)=2(b-a)/2=b-a=L(\chi_A)\). For \(A=\{1\}\), the inverse image within \([0,1/2]\) is the singleton \(\{1/2\}\), so both sides are zero. The identity holds on the algebra and therefore, by the theorem, for all bounded Borel-measurable \(f\).
Check: for \(f(y)=y^2\), the left side is \(\int_0^1y^2\,dy=1/3\). The right side is \(2\int_0^{1/2}(2x)^2\,dx=8[x^3/3]_0^{1/2}=8(1/24)=1/3\).
Worked Example: Functions Constant on Two Blocks
Let \(X=\{1,2,3,4\}\), with \(\mathcal{M}\) its power set, and let \(B_1=\{1,2\}\), \(B_2=\{3,4\}\). Take the algebra \(\mathcal{A}=\{\varnothing,X,B_1,B_2\}\). Define probability measures \(\mu\) and \(\nu\) by the atom weights \(\mu=(1/4,1/4,1/2,0)\) and \(\nu=(0,1/2,1/4,1/4)\), in the order \(1,2,3,4\). Each block has measure \(1/2\) under either measure, so their integral functionals agree on indicators of \(\mathcal{A}\).
The functionals are positive and linear. Since \(X\) is finite, if \(0\le f_n\uparrow f\), then each of the four coordinate sequences increases to its limit, and the finite weighted sums defining both integrals converge to their values at \(f\). Thus the bounded increasing-limit condition holds. The theorem implies equality of the integrals for every bounded \(\sigma(\mathcal{A})\)-measurable function, which here means every function constant on each block. For \(f=3\chi_{B_1}-2\chi_{B_2}\), both integrals equal \(3(1/2)-2(1/2)=1/2\).
Check: summing atom by atom gives \(\int f\,d\mu=3(1/4)+3(1/4)-2(1/2)-2(0)=1/2\), and \(\int f\,d\nu=3(0)+3(1/2)-2(1/4)-2(1/4)=1/2\).
Why the Limit Condition Matters
A common mistake is to check an identity on indicators and then pass to a pointwise limit without checking monotonicity and boundedness. For example, on \([0,1]\) let \(f_n(x)=n\chi_{(0,1/n)}(x)\). For every \(x\), \(f_n(x)\to0\), but \(\int_0^1 f_n(x)\,dx=n(1/n)=1\) for every \(n\), whereas \(\int_0^1\lim_n f_n(x)\,dx=0\). This sequence is not increasing and has no common finite bound. The bounded monotone class argument does not permit this limit passage. Its force comes precisely from the verified increasing-limit continuity, not from pointwise convergence alone.
The method is useful whenever a bounded-function identity has a simple verification on a generating algebra and both sides have the needed continuity. The algebra supplies the initial tests; the monotone class theorem supplies all generated measurable sets; linearity and bounded increasing approximation then supply all bounded measurable functions in the generated sigma-algebra. Keeping these steps separate makes the hypotheses visible and prevents an unjustified interchange of a limit with an integral.
Check Your Understanding
For each question, identify which part of the argument supplies the needed conclusion.
- Why does equality on indicators of an algebra imply equality on indicators of every set in its generated sigma-algebra?
- Which continuity property of an integral against a finite measure is used for a bounded increasing sequence of nonnegative functions?
- In the reflection example, what is the inverse image of \([a,b)\) under \(x\mapsto1-x\), and what is its measure?
- For the two-block example, describe exactly which functions are measurable with respect to \(\sigma(\mathcal{A})\).
- Which hypotheses fail for \(f_n=n\chi_{(0,1/n)}\) when one tries to pass an integral identity through its pointwise limit?