# Sharp assumption for preserving Lebesgue measurability by left composition

Let $$g: [0, 1] \to \mathbb R$$ be a Lebesgue-measurable function (in the classical sense: the inverse images of Borel sets are Lebesgue-measurable). It is a classical fact in analysis that $$f \circ g$$ is Lebesgue-measurable as soon as $$f$$ is continuous, for instance, or Borel-measurable (the inverse images of Borel sets are Borel), but not necessarily if $$f$$ is only Lebesgue-measurable. My question is: what is the sharpest assumption that one can put on $$f$$ guaranteeing that $$f \circ g$$ is Lebesgue-measurable for every Lebesgue-measurable $$g$$?

More precisely, consider the class \begin{aligned} \mathcal F = \{f: \mathbb R \to \mathbb R \mid {} & g: [0, 1] \to \mathbb R \text{ is Lebesgue-measurable} \implies \\ & f \circ g \text{ is Lebesgue-measurable}\}. \end{aligned} $$\mathcal F$$ contains all Borel-measurable functions, but does it contain other functions? Or is it equal to the set of all Borel-measurable functions?

This question looks quite natural and I imagined there should be some classical result in real analysis providing its answer, but I could not find it in standard textbooks. I initially conjectured that $$\mathcal F$$ coincides with the set of all Borel-measurable functions. This would mean that, if $$f$$ is not a Borel-measurable function, then there exists a Lebesgue-measurable $$g$$ such that $$f \circ g$$ is not Lebesgue-measurable. The idea would be to pick such an $$f$$, pick a Borel set $$A$$ such that $$B = f^{-1}(A)$$ is not Borel, and try to construct a Lebesgue-measurable $$g$$ such that $$g^{-1}(B)$$ is not Lebesgue-measurable, but I cannot see how to construct such a $$g$$ while keeping its Lebesgue-measurability. Any ideas or any references to this question?

The answer is: $$\cal F$$ is the family of universally measurable functions.

For simplicity, let us consider functions on $$[0,1]$$ rather than on $$\mathbb R$$. Let $$\cal B$$ be the family of Borel sets, $$\cal B^\star$$ the family of universally measurable sets, and $$\cal L$$ the family of Lebesgue sets.

Clearly, it is sufficient that $$f$$ is universally measurable: every $$\cal B/\cal L$$-measurable function is in fact $$\cal B^\star/\cal L$$ measurable (for clearly $$\cal L^\star = \cal L$$); see, for example, the PlanetMath entry.

The above condition turns out to be necessary, too. Indeed: suppose that $$f$$ is not universally measurable, that is, $$A := f^{-1}(B) \notin \cal B^\star$$ for some $$B \in \cal B$$. We will construct a continuous (!) function $$g$$ such that $$g^{-1}(A)$$ is not in $$\cal L$$. Of course, this implies that $$f \notin \cal F$$.

There is a Borel probability measure $$\mu$$ such that $$A$$ is not $$\mu$$-measurable. Let $$\lambda$$ be the Lebesgue measure on $$[0, 1]$$. Considering $$\tfrac{1}{2} \mu + \tfrac{1}{2} \lambda$$ rather than $$\mu$$, we may assume that the distribution function $$h$$ of $$\mu$$ is strictly increasing. Removing the atoms of $$\mu$$ and renormalising it, we can make $$\mu$$ atomless and $$h$$ is continuous.

We have $$\mu(E) = \lambda(h(E))$$ for every Borel set $$E$$. If $$h(A)$$ were Lebesgue measurable, we would have two Borel sets $$F_1$$ and $$F_2$$ such that $$F_1 \subseteq h(A) \subseteq F_2$$ and $$\lambda(F_2 \setminus F_1) = 0$$. But then $$E_1 = h^{-1}(F_1)$$ and $$E_2 = h^{-1}(F_2)$$ would be Borel sets such that $$E_1 \subseteq A \subseteq E_2$$ and $$\mu(E_2 \setminus E_1) = \lambda(h(E_2 \setminus E_1)) = \lambda(F_2 \setminus F_1) = 0 ,$$ and consequently $$A$$ would be $$\mu$$-measurable.

Now if $$g$$ is the inverse of $$h$$, then $$g$$ is continuous and strictly increasing, and $$g^{-1}(A) = h(A)$$ is not Lebesgue-measurable.

• Thanks for the excellent reply! That's a very nice construction! – Guilherme Mazanti Jul 30 '20 at 17:21

Take any $$f\in\mathcal F$$. Take any real $$b$$ and any real $$a>0$$, and let $$g(x):=ax+b$$ for $$x\in[0,1]$$. Then the function $$g\colon[0,1]\to\mathbb R$$ is Borel-measurable and hence the function $$h:=f\circ g$$ is Borel-measurable. So, for any Borel set $$A\subseteq\mathbb R$$, the set $$f^{-1}(A)\cap[b,a+b]=ah^{-1}(A)+b:=\{ax+b\colon x\in h^{-1}(A)\}$$ is Borel, for any interval $$[b,a+b]$$, which implies that the set $$f^{-1}(A)$$ is Borel. So, $$f$$ is Borel-measurable.

Thus, $$\mathcal F$$ coincides with the set of all Borel-measurable from $$\mathbb R$$ to $$\mathbb R$$.

• I now realize my question is not necessarily clear enough: I use two notions of measurability, Lebesgue and Borel. Whenever I said only "measurable", I mean Lebesgue-measurable. I will edit the question to make this more precise. – Guilherme Mazanti Jul 30 '20 at 16:08