Why Composition Can Fail
The composition of two measurable functions is measurable when the inner function takes values in the measurable space on which the outer function is measurable. The qualification matters: a function that is Lebesgue measurable on \(\mathbb{R}\) need not be measurable for the Borel sigma-algebra on \(\mathbb{R}\). A Borel-measurable inner map only guarantees that preimages of Borel sets are measurable in its domain; it gives no such guarantee for preimages of arbitrary Lebesgue-measurable sets.
We will construct a real-valued Lebesgue-measurable function \(f\) on \((\mathbb{R},\mathcal L,m)\) that is not Borel measurable, and a Borel-measurable function \(g\) on the Borel domain \(([0,1],\mathcal B_{[0,1]},m)\), such that \(f\circ g\) is not Borel measurable on that domain. Here \(\mathcal B_{[0,1]}\) is the Borel sigma-algebra relative to \([0,1]\), and \(m\) on this Borel space is finite. We will also check that the same composition is measurable if the domain is instead given its completed Lebesgue sigma-algebra.
A Lebesgue-Measurable Set That Is Not Borel
Let \(C\) be the middle-thirds Cantor set. It is the intersection of the finite-stage sets \(C_n\), where \(C_n\) is the union of \(2^n\) closed intervals, each of length \(3^{-n}\). Thus \(C\subseteq C_n\) and \(m(C_n)=(2/3)^n\), so \(0\le m(C)\le (2/3)^n\) for every \(n\). Since \((2/3)^n\to0\), \(m(C)=0\).
The Cantor set has cardinality \(\mathfrak c\), the cardinality of \(\mathbb R\). In fact, the map from binary sequences \((\varepsilon_k)_{k\ge1}\), with each \(\varepsilon_k\in\{0,1\}\), to \(\sum_{k\ge1}2\varepsilon_k3^{-k}\) maps into \(C\) and is injective. To check injectivity, if two sequences first differ at index \(k\), the contribution \(2\cdot3^{-k}\) at that index exceeds the largest possible opposing tail, \(\sum_{j>k}2\cdot3^{-j}=3^{-k}\). Also \(C\subseteq\mathbb R\), so its cardinality is at most \(\mathfrak c\).
Proof. Since \(m(C)=0\), every subset \(A\subseteq C\) is Lebesgue measurable and has measure zero. Indeed, its outer measure is zero; for every \(E\subseteq\mathbb R\), subadditivity gives \(m^*(E)\le m^*(E\cap A)+m^*(E\setminus A)\), while \(m^*(E\cap A)=0\) and \(m^*(E\setminus A)\le m^*(E)\). Hence equality holds in the Carathéodory criterion, so \(A\in\mathcal L\). This is the completeness of Lebesgue measure in this instance. There are \(2^{\mathfrak c}\) subsets of \(C\). There are at most \(\mathfrak c\) Borel subsets of \(\mathbb R\): using a countable base, Borel sets can be coded by countable well-founded trees whose nodes carry countable-base labels and whose operations are countable unions and complements. There are at most \(\mathfrak c\) such countable codes. Cantor’s theorem gives \(2^{\mathfrak c}>\mathfrak c\), so not every subset of \(C\) is Borel. Choose one such non-Borel subset and call it \(A\). The first paragraph shows that \(A\) is Lebesgue measurable. \(\square\)
This existence argument is enough for the counterexample; no explicit description of the individual points of \(A\) is needed. Since \(A\) is not Borel, it is nonempty, and we may also choose a point \(a\in A\) when needed below.
The Composition Rule and Its Missing Hypothesis
Proof. Fix \(D\in\mathcal U\). By measurability of \(f\), \(f^{-1}(D)\in\mathcal T\). By measurability of \(g\), \(g^{-1}(f^{-1}(D))\in\mathcal S\). Since \(g^{-1}(f^{-1}(D))=(f\circ g)^{-1}(D)\), the preimage of every \(D\in\mathcal U\) under \(f\circ g\) belongs to \(\mathcal S\). This is the definition of measurability of the composition. \(\square\)
For Borel measurability, the theorem asks that \(f^{-1}(D)\) be Borel whenever \(D\) is Borel. Lebesgue measurability of \(f\) only says that these preimages belong to \(\mathcal L\). A Lebesgue-measurable preimage need not be Borel, which is exactly where the composition argument can stop.
Worked Example: The Identity Map Exposes the Failure
Use the non-Borel Lebesgue-measurable set \(A\subseteq C\) above. Define \(f:\mathbb R\to\mathbb R\) and \(g:[0,1]\to\mathbb R\) by \(f=\chi_A\) and \(g(x)=x\). The outer map \(f\) is real-valued and Lebesgue measurable on \((\mathbb R,\mathcal L,m)\): for any open \(U\subseteq\mathbb R\), its preimage is one of \(\varnothing,A,A^c,\mathbb R\), all of which are in \(\mathcal L\). It is not Borel measurable, since \(\{y:f(y)>1/2\}=A\) is not Borel.
The inner map \(g\) is Borel measurable from \(([0,1],\mathcal B_{[0,1]},m)\) to \((\mathbb R,\mathcal B_{\mathbb R},m)\): for any Borel \(D\subseteq\mathbb R\), \(g^{-1}(D)=D\cap[0,1]\), which is relatively Borel. But \(f\circ g=\chi_A\) on \([0,1]\), and its strict level set at \(1/2\) is
Thus \(f\circ g\) is not Borel measurable on the Borel domain: the displayed level set is not in \(\mathcal B_{[0,1]}\). It is, however, Lebesgue measurable on \(([0,1],\mathcal L|_{[0,1]},m)\), because \(A\in\mathcal L\). The failure is specifically Borel measurability on the stated domain, not Lebesgue measurability after completing the domain sigma-algebra.
Check: the composition takes only the values \(0\) and \(1\); its strict \(1/2\)-level set is exactly \(A\), not a larger or smaller set. Since \(A\) is Lebesgue measurable and non-Borel, these two measurability conclusions differ as claimed.
Two Variations That Clarify the Example
Worked Example: A Non-Borel Outer Function Can Still Give a Measurable Composition
Keep \(f=\chi_A\), choose \(a\in A\), and define \(g:[0,1]\to\mathbb R\) by \(g(x)=a\). This constant function is Borel measurable: for each Borel \(D\subseteq\mathbb R\), its preimage is \([0,1]\) if \(a\in D\), and \(\varnothing\) otherwise. Since \(a\in A\), \((f\circ g)(x)=f(a)=1\) for every \(x\in[0,1]\). Every level-set preimage of this constant function is either \([0,1]\) or \(\varnothing\), so the composition is Borel measurable.
Check: for the strict level set \(\{x:(f\circ g)(x)>1/2\}\), every \(x\in[0,1]\) qualifies, giving \([0,1]\), a Borel set. The outer function’s failure of Borel measurability does not force every composition with it to fail.
Worked Example: Changing the Inner Map Preserves the Obstruction
Again let \(f=\chi_A\), but take the Borel domain \(([0,1/2],\mathcal B_{[0,1/2]},m)\) and define \(g(x)=2x\). For any Borel \(D\subseteq\mathbb R\), \(g^{-1}(D)=\{x\in[0,1/2]:2x\in D\}\) is relatively Borel, since \(x\mapsto2x\) is continuous. The strict level set is
Call the final set \(A/2\). It is not Borel: if it were, its image under the continuous map \(x\mapsto2x\) would be Borel, contradicting that \(2(A/2)=A\) is not Borel. It is Lebesgue measurable and null because it is a subset of the scaled null set \(C/2\). Thus this composition, too, is not Borel measurable on its Borel domain, although it is Lebesgue measurable if the domain is completed.
Check: each \(a\in A\subseteq[0,1]\) gives \(a/2\in[0,1/2]\), so no points are lost through the domain restriction. Conversely, \(2x\in A\) implies \(x=a/2\) for the point \(a=2x\in A\), proving the set equality in both directions.
A Common Mistake
A tempting but invalid calculation is: “For Borel \(D\), \(f^{-1}(D)\) is measurable; therefore \(g^{-1}(f^{-1}(D))\) is measurable.” The word “measurable” changes meaning in the middle. For our \(f=\chi_A\), take the Borel target set \(D=(1/2,\infty)\). Then \(f^{-1}(D)=A\), which is Lebesgue measurable but not Borel. The Borel measurability of \(g\) only controls \(g^{-1}(E)\) for Borel \(E\subseteq\mathbb R\); it does not give a Borel preimage for this \(A\). With \(g\) the identity, \(g^{-1}(A)=A\), which is indeed not Borel.
The correction is to name the sigma-algebras each time: the usual composition theorem applies when the outer function is measurable for the target sigma-algebra used by the inner function. For Borel-measurable composition, both maps must be Borel measurable. Alternatively, one can work with a completed domain and verify separately that the relevant preimages are Lebesgue measurable; the identity example has that property, but Borel measurability alone does not assert it.
Check Your Understanding
In each question, specify the domain sigma-algebra as well as the set whose measurability is being tested.
- Why is every subset of the Cantor set Lebesgue measurable, and why does that fact not imply every such subset is Borel?
- For \(f=\chi_A\) and \(g(x)=x\) on \([0,1]\), compute the strict level set \(\{x:(f\circ g)(x)>1/2\}\).
- State the sigma-algebra hypothesis on \(f\) that makes the measurable-composition theorem apply to two Borel-measurable maps.
- For the map \(g(x)=2x\) on \([0,1/2]\), verify both inclusions in the equality between the strict \(1/2\)-level set and \(A/2\).
- Does the identity-map example give a Lebesgue-measurable composition on the completed domain \(([0,1],\mathcal L|_{[0,1]},m)\)? Explain using the set \(A\).