Why the Outer Map Must Be Borel-Measurable
Measurability is defined through preimages: a function is measurable when preimages of measurable sets belong to the sigma-algebra on its domain. Composition is therefore governed by a simple identity: the preimage under a composition is a preimage under the inner map of a preimage under the outer map. The crucial point is that the inner function is assumed measurable for Borel subsets of the real line. Thus the outer map must take Borel sets back to Borel sets.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed here. A real-valued function \(f:X\to\mathbb{R}\) is \(\mathcal{M}\)-measurable if \(f^{-1}(C)\in\mathcal{M}\) for every \(C\in\mathcal{B}_{\mathbb{R}}\). A function \(g:\mathbb{R}\to\mathbb{R}\) is Borel-measurable if \(g^{-1}(B)\in\mathcal{B}_{\mathbb{R}}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\). These definitions concern sets and hold everywhere; they do not mean that functions are identified up to equality \(\mu\)-a.e.
Proof. To prove measurability, take any \(B\in\mathcal{B}_{\mathbb{R}}\). The definition of composition gives \((g\circ f)^{-1}(B)=f^{-1}(g^{-1}(B))\). Since \(g\) is Borel-measurable, \(g^{-1}(B)\in\mathcal{B}_{\mathbb{R}}\). Since \(f\) is \(\mathcal{M}\)-measurable, the preimage of this Borel set belongs to \(\mathcal{M}\). Hence \((g\circ f)^{-1}(B)\in\mathcal{M}\) for every Borel \(B\), as required. \(\square\)
The theorem does not require \(g\) to be continuous. Earlier material on continuous functions gives one useful way to produce Borel-measurable maps, since every continuous map between the real line and itself is Borel-measurable. The composition theorem applies more broadly, including to discontinuous outer maps.
Worked Example: Composing with an Indicator of the Rationals
Let \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and \(f(x)=x\). Define \(g(t)=1\) for \(t\in\mathbb{Q}\) and \(g(t)=0\) for \(t\notin\mathbb{Q}\). Verify that \(g\circ f\) is Borel-measurable.
The set \(\mathbb{Q}\) is countable, hence Borel, and its complement is Borel. For any open set \(U\subseteq\mathbb{R}\), there are four possibilities. If \(U\) contains neither \(0\) nor \(1\), then \(g^{-1}(U)=\varnothing\). If \(U\) contains \(0\) but not \(1\), then \(g^{-1}(U)=\mathbb{R}\setminus\mathbb{Q}\). If \(U\) contains \(1\) but not \(0\), then \(g^{-1}(U)=\mathbb{Q}\). If it contains both, then \(g^{-1}(U)=\mathbb{R}\). Each of these sets is Borel; in fact, this verifies the open-set test for \(g\).
The identity \(f\) is Borel-measurable, so the theorem applies. Directly, \((g\circ f)(x)=1\) on \(\mathbb{Q}\) and \(0\) on its complement. Its preimage of an open \(U\) is one of the same four sets just listed, so it is Borel. Check: for \(U=(1/2,3/2)\), the preimage is \(\mathbb{Q}\), which is Borel; for \(U=(-1/2,1/2)\), it is \(\mathbb{R}\setminus\mathbb{Q}\), also Borel.
Outer Maps Defined Only on a Borel Domain
Some familiar functions are not defined on all of \(\mathbb{R}\). The appropriate version of the theorem uses the relative Borel sigma-algebra on their domain. If \(D\subseteq\mathbb{R}\), its relative Borel sigma-algebra is \(\mathcal{B}_{D}=\{D\cap C:C\in\mathcal{B}_{\mathbb{R}}\}\). A map \(g:D\to\mathbb{R}\) is Borel-measurable if \(g^{-1}(B)\in\mathcal{B}_D\) for every \(B\in\mathcal{B}_{\mathbb{R}}\).
Proof. Fix \(B\in\mathcal{B}_{\mathbb{R}}\). By relative measurability there is \(C\in\mathcal{B}_{\mathbb{R}}\) such that \(g^{-1}(B)=D\cap C\). Since every value of \(f\) lies in \(D\), we have \((g\circ f)^{-1}(B)=f^{-1}(D\cap C)=f^{-1}(C)\). The last set is in \(\mathcal{M}\) because \(f\) is measurable. This holds for every Borel \(B\), proving the claim. \(\square\)
The range condition is pointwise: \(f(x)\in D\) for every \(x\), so the composition is defined everywhere on \(X\). If the condition holds only almost everywhere, then \(g\circ f\) may not be defined on the exceptional set; a separate definition there would be needed before asserting measurability on all of \(X\).
Worked Example: The Logarithm on Positive Inputs
Let \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and \(f(x)=x\) on \(X_+=(0,\infty)\), viewed as the measurable space \((X_+,\mathcal{B}_{X_+})\). Show that \(x\mapsto\ln x\) is measurable on \(X_+\), and compute a preimage explicitly.
The set \(D=(0,\infty)\) is Borel. The logarithm \(\ln:D\to\mathbb{R}\) is continuous, hence Borel-measurable on \(D\). The identity \(f:X_+\to D\) is measurable and takes all its values in \(D\). The domain-restricted composition theorem therefore gives measurability of \(x\mapsto\ln x\) on \(X_+\).
For a direct set calculation, let \(a<b\) be real numbers. Since the logarithm is strictly increasing and its inverse is the exponential function, \((\ln)^{-1}((a,b))=(e^a,e^b)\). For example, with \(a=0\) and \(b=\ln 4\), this preimage is \((1,4)\), because \(e^0=1\) and \(e^{\ln 4}=4\). This is Borel in \(X_+\). Check: every \(x\in(1,4)\) satisfies \(0<\ln x<\ln 4\), and if \(0<\ln x<\ln 4\), exponentiating gives \(1<x<4\).
The same reasoning applies to other domain-restricted continuous maps. For example, \(t\mapsto\sqrt{t}\) is continuous on the Borel domain \([0,\infty)\); if a measurable real-valued \(f\) takes values in \([0,\infty)\) everywhere, then \(x\mapsto\sqrt{f(x)}\) is measurable. The domain condition matters: if \(f\) has negative values, this real-valued composition is not defined there.
Worked Example: Square Root on a Restricted Range
Let \(X=\mathbb{R}\) with its Borel sigma-algebra, and define \(f(x)=x^2\). Verify measurability of \(x\mapsto\sqrt{f(x)}\) using the restricted-domain theorem, and check the result explicitly.
The function \(f\) is continuous, hence Borel-measurable, and \(f(x)\in[0,\infty)\) for every \(x\). The square-root map is continuous on \(D=[0,\infty)\), so it is Borel-measurable there. Thus the composition is measurable. Its value is \(\sqrt{x^2}=|x|\): if \(x\ge0\), the nonnegative square root of \(x^2\) is \(x\); if \(x<0\), it is \(-x\). For any \(a\ge0\), the preimage of \([0,a)\) under this composition is \((-a,a)\) when \(a>0\), and is empty when \(a=0\). These sets are Borel.
Check: with \(a=3\), the preimage of \([0,3)\) is \((-3,3)\); every \(x\) in this interval has \(|x|<3\), and every \(x\) with \(|x|<3\) lies in the interval. This agrees with the formula \(\sqrt{x^2}=|x|\).
A Common Mistake: Replacing Borel Measurability by Lebesgue Measurability
The theorem cannot in general be weakened by asking only that \(g\) be Lebesgue-measurable. The inner function \(f\) pulls back Borel sets, not arbitrary Lebesgue-measurable sets. Here is a concrete setting where the distinction matters.
Take \(X=\mathbb{R}\) with \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and let \(f(x)=x\), which is measurable on this space. Let \(K\) be the Cantor set. It is a Borel null set and has cardinality equal to that of \(\mathbb{R}\). There are \(2^{|\mathbb{R}|}\) subsets of \(K\), while there are only \(|\mathbb{R}|\) Borel subsets of \(\mathbb{R}\); hence some subset \(A\subseteq K\) is not Borel. Since Lebesgue measure is complete, every subset of the null set \(K\) is Lebesgue-measurable. Define \(g=\chi_A:\mathbb{R}\to\mathbb{R}\). This \(g\) is Lebesgue-measurable, but it is not Borel-measurable: \(g^{-1}((1/2,3/2))=A\), which is not Borel.
Now \(g\circ f=\chi_A\) on the Borel measurable space \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\). Its preimage of \((1/2,3/2)\) is \(A\notin\mathcal{B}_{\mathbb{R}}\), so \(g\circ f\) is not measurable on that space. The example does not contradict composition with Borel outer maps: \(g\) fails precisely the Borel hypothesis.
Check Your Understanding
For each question, identify the relevant domain and sigma-algebra before applying a composition result.
- Let \(f:X\to\mathbb{R}\) be measurable and \(g(t)=\chi_{[2,\infty)}(t)\). Explain why \(g\circ f\) is measurable, and express its set of points where the value is \(1\).
- Suppose \(f:X\to\mathbb{R}\) is measurable and \(f(x)>0\) for every \(x\). State precisely why \(x\mapsto\ln(f(x))\) is measurable.
- Let \(D=[-1,1]\) and let \(g:D\to\mathbb{R}\) be Borel-measurable. What additional condition on \(f\) lets you conclude that \(g\circ f\) is defined and measurable everywhere?
- For \(f(x)=x^2\) on \(\mathbb{R}\), compute the preimage of \([0,2)\) under \(x\mapsto\sqrt{f(x)}\).
- In the counterexample above, which preimage shows that \(g\) is not Borel-measurable, and why does that prevent the composition theorem from applying?