Measure-Theoretic Proof Workshop

Two Countable Constructions

The previous tutorial used rational boxes to prove that measurable coordinate functions form a measurable pair, then applied a continuous operation to that pair. This tutorial develops two further proof techniques: defining a function separately on countably many measurable pieces, and taking the pointwise supremum of countably many measurable functions. In both arguments, the key step is to express a preimage or a level set using countable unions and intersections.

Throughout, \((X,\mathcal{F})\) is a measurable space, and real-valued functions are measurable with respect to the Borel sigma-algebra on \(\mathbb{R}\). We will use the Threshold Characterization of Measurability: a real-valued function is measurable if and only if each of its strict superlevel sets is measurable. We will also use the closure of a sigma-algebra under countable unions and intersections.

Measurable Functions Defined Piece by Piece

Suppose a measurable space is divided into countably many measurable regions, and on each region we use a different measurable function. To verify the resulting function, fix a Borel set \(B\) in the range. A point belongs to the preimage of \(B\) precisely when it lies in one of the regions and the function assigned to that region takes a value in \(B\). This describes the preimage as a countable union of measurable sets.

Definition (Countable Measurable Partition): A sequence \((E_n)_{n=1}^{\infty}\) is a countable measurable partition of \(X\) if each \(E_n\in\mathcal{F}\), the sets are pairwise disjoint, and \(\bigcup_{n=1}^{\infty}E_n=X\).

Theorem (Countable Pasting of Measurable Functions): Let \((E_n)_{n=1}^{\infty}\) be a countable measurable partition of \(X\), and let \(f_n:X\to\mathbb{R}\) be measurable for every \(n\). Define \(h:X\to\mathbb{R}\) by \(h(x)=f_n(x)\) when \(x\in E_n\). Then \(h\) is measurable.

Proof. The partition property ensures that every \(x\in X\) lies in exactly one \(E_n\), so the definition gives exactly one value \(h(x)\). Let \(B\in\mathcal{B}(\mathbb{R})\). For each \(n\), the set \(f_n^{-1}(B)\) belongs to \(\mathcal{F}\), and so does \(E_n\cap f_n^{-1}(B)\). By the definition of \(h\),

h^{-1}(B) = \bigcup_{n=1}^{\infty}\bigl(E_n\cap f_n^{-1}(B)\bigr).

The right-hand side is a countable union of measurable sets, hence belongs to \(\mathcal{F}\). This holds for every Borel set \(B\), so \(h\) is measurable. \(\square\)

A useful variation allows the pieces to cover only a measurable set \(E\subseteq X\). Define the function by the indicated formulas on a countable measurable partition of \(E\), and assign a fixed real value \(c\) on \(X\setminus E\). This is a countable pasting on \(X\) after adjoining \(X\setminus E\) as one more piece and using the constant function \(x\mapsto c\) there. The constant function is measurable because the preimage of any Borel set is either \(X\) or \(\varnothing\).

Worked Example: A Function Defined on Alternating Intervals

Let \(E_n=[n,n+1)\) for integers \(n\geq 0\), and let \(E_{-}=(-\infty,0)\). These Borel sets are pairwise disjoint and cover \(\mathbb{R}\). Define \(h:\mathbb{R}\to\mathbb{R}\) by

h(x)= \begin{cases} x^2,&x\in E_{-},\\ (-1)^n x,&x\in E_n,\quad n\geq 0. \end{cases}

On \(E_{-}\), the assigned function \(x\mapsto x^2\) is continuous. On each \(E_n\), the assigned function \(x\mapsto(-1)^n x\) is continuous. The countable pasting theorem applies after enumerating the pieces \(E_{-},E_0,E_1,\ldots\); thus \(h\) is measurable. For example, \(h(-2)=(-2)^2=4\), \(h(0.4)=0.4\), and \(h(1.4)=-1.4\), since \(0.4\in E_0\) and \(1.4\in E_1\). The endpoint convention matters: \(1\in E_1\), not \(E_0\), so \(h(1)=-1\).

The partition hypotheses prevent ambiguity. If the pieces overlap and the assigned formulas disagree there, the phrase “the function defined by the pieces” may not define a function at all. If they do not cover \(X\), a value must be supplied on the uncovered set. In either situation, checking these details comes before applying the theorem.

Pointwise Suprema and Level Sets

For a sequence of real-valued functions, the pointwise supremum may be \(+\infty\) at some points, even though every individual function is finite-valued. We therefore first state the result using strict superlevel sets, which makes sense for an extended-real-valued function. If the supremum is finite everywhere, the Threshold Characterization then gives ordinary real-valued measurability.

Definition (Measurable Extended-Real-Valued Function): A function \(g:X\to[-\infty,+\infty]\) is measurable if \(\{x\in X:g(x)>a\}\in\mathcal{F}\) for every \(a\in\mathbb{R}\).

Theorem (Measurability of a Countable Pointwise Supremum): Let \(f_n:X\to\mathbb{R}\) be measurable for every positive integer \(n\), and define \(g(x)=\sup_{n\geq 1}f_n(x)\), allowing \(g(x)=+\infty\). Then \(g\) is measurable as an extended-real-valued function. If \(g(x)\) is finite for every \(x\), then \(g:X\to\mathbb{R}\) is measurable.

Proof. Fix \(a\in\mathbb{R}\). At a point \(x\), the supremum of the numbers \(f_n(x)\) is strictly greater than \(a\) if and only if at least one of those numbers is strictly greater than \(a\). Indeed, if every \(f_n(x)\leq a\), then \(a\) is an upper bound and \(g(x)\leq a\). Conversely, if \(f_n(x)>a\) for some \(n\), then \(g(x)\geq f_n(x)>a\). Therefore

\{x:g(x)>a\} = \bigcup_{n=1}^{\infty}\{x:f_n(x)>a\}.

Each set on the right belongs to \(\mathcal{F}\), since \(f_n\) is measurable. The union is countable, so it belongs to \(\mathcal{F}\). This holds for every real \(a\), proving extended-real-valued measurability by definition. If \(g\) is finite everywhere, it is real-valued, and its measurable strict superlevel sets imply measurability by the Threshold Characterization. \(\square\)

Worked Example: A Supremum That Is Never Attained

For each positive integer \(n\), define \(f_n:\mathbb{R}\to\mathbb{R}\) by \(f_n(x)=x-\frac{1}{n}\). Each \(f_n\) is continuous and hence measurable. At any fixed \(x\), every term satisfies \(f_n(x)<x\), so the supremum is at most \(x\). On the other hand, for every \(\varepsilon>0\), choose \(n\) so large that \(1/n<\varepsilon\). Then \(f_n(x)=x-1/n>x-\varepsilon\). Thus no number strictly below \(x\) is an upper bound, and

\sup_{n\geq 1}f_n(x)=x.

The supremum is finite and measurable, but it is not attained: for every \(n\), \(x-1/n<x\). This distinction is useful whenever a proof uses a supremum; one must not assume a maximizing index exists.

Worked Example: A Measurable Envelope of Continuous Functions

Define \(f_n:\mathbb{R}\to\mathbb{R}\) by \(f_n(x)=-|x-n|\), for \(n\geq1\). Each function is continuous. Their pointwise supremum is finite: every \(f_n(x)\leq0\), while \(f_1(x)=-|x-1|\) is a finite lower bound for the supremum at each fixed \(x\). Thus \(g(x)=\sup_{n\geq1}f_n(x)\) is real-valued and measurable by the theorem. For \(x=3\), the first few values are \(-2,-1,0,-1\), corresponding to \(n=1,2,3,4\), so \(g(3)=0\). For \(x=3.4\), \(f_3(3.4)=-0.4\) and \(f_4(3.4)=-0.6\); every other integer \(n\) is at least as far from \(3.4\) as \(3\) or \(4\), so \(g(3.4)=-0.4\). The theorem establishes measurability on the whole real line without needing to identify the envelope’s formula everywhere.

Worked Example: A Countable Pasting with a Default Value

Let \(A_n=[n,n+1)\) for \(n\geq1\), and let \(A=\bigcup_{n=1}^{\infty}A_n=[1,\infty)\). Define \(u:\mathbb{R}\to\mathbb{R}\) by \(u(x)=1/n\) for \(x\in A_n\), and \(u(x)=0\) for \(x\notin A\). The sets \(A_n\) and \(\mathbb{R}\setminus A\) form a countable measurable partition. On each \(A_n\), use the constant function \(x\mapsto1/n\); on the complement, use the constant function \(x\mapsto0\). The countable pasting theorem proves that \(u\) is measurable. In particular, \(u(1.2)=1\), \(u(2.7)=1/2\), and \(u(0)=0\). The complement is part of the partition, so the definition assigns a value at every real number.

Why Countability Is Essential

The proofs above work because sigma-algebras are closed under countable unions, not arbitrary unions. It is not valid to replace “countable” by “any collection” in either theorem without additional assumptions. For a concrete warning, let \(X=\mathbb{R}\) with the countable-cocountable sigma-algebra, consisting of the countable sets and the sets whose complements are countable. Each singleton \(\{x\}\) is measurable. For each \(x\in(0,1)\), the indicator function \(\mathbf{1}_{\{x\}}\) is therefore measurable. But the pointwise supremum over this uncountable family is \(\mathbf{1}_{(0,1)}\). The interval \((0,1)\) is uncountable and its complement in \(\mathbb{R}\) is uncountable, so it is not measurable in this sigma-algebra. Thus an uncountable supremum of measurable functions need not be measurable.

A dependable proof strategy is to begin with a measurable target set or a threshold, translate membership into conditions on the pieces or component functions, and then check that the resulting expression uses only countable sigma-algebra operations. The countable pasting theorem uses intersections with the pieces followed by a union. The supremum theorem uses a union of strict superlevel sets. In both cases, the identity describing the set is the central part of the proof; the closure properties finish it.

Key takeaway: Countable constructions preserve measurability when their relevant preimages or level sets can be written as countable unions and intersections of measurable sets. Check coverage, uniqueness, and whether a supremum is finite before applying the corresponding real-valued conclusion.

Check Your Understanding

Use the set identities and hypotheses in the two theorems to answer the following questions.

Why must the pieces in a measurable partition cover \(X\) and be pairwise disjoint before the pasted formula defines a function everywhere?
Write the preimage \(h^{-1}(B)\) for a countable pasting in terms of the partition pieces and the functions \(f_n\).
Why does \(\sup_n f_n(x)>a\) imply that \(f_n(x)>a\) for at least one index \(n\)?
What additional condition lets the extended-real-valued supremum theorem give a real-valued measurable function?
Which closure property fails to justify arbitrary, uncountable unions of measurable sets?