The class of functions preserving measurability of functions in post composition

Question

Is it possible to characterize the class of functions $g:{\bf R}\rightarrow {\bf R}$ which satisfies the following condition: for every measurable function $u:{\bf R}\rightarrow {\bf R}$ it follows that $g\circ u$ is a measurable function? Let us say that such functions $g$ are characterised by the property (L). So, the question is: what the property (L) exactly is?

${\bf DISCUSSSION.}$

Remark(1). Here measurability is meant in the sense of Lebesgue (so it is Lebesgue measurability): more precisely: We say that $u:{\bf R}\rightarrow {\bf R}$ is measurable iff for every open set $U\subseteq {\bf R}$ we have that $u^{\leftarrow}(U)$ is a (Lebesgue) measurable set.

Remark(2). I guess, the same question can be asked if we replace the notion of Lebesgue measurability by the notion of Borel measurability. Let us say the aforementioned property of function $g$ related to Borel measurability is (B) property. Can we establish which property exactly (B) property is?

Remark(3). The question is inspired by two superposition theorems in textbook G. Leoni: A First Course in Sobolev Spaces. 2nd edition, 2017, where, in chapter 2 (chapter 3, resp.), in Theorem 2.55, page 55 (Theorem 3.55, page 92, resp.), is established that the class of functions $g$ preserving $BPV_{loc}$-property ($AC_{loc}$-property, resp.) in post composition is exactly the class of locally Lipschitz functions.

Remark(4). Here are some questions which are generally related to this question (but which consider a different problem):

Composition of measurable & continuous functions, is it measurable?

Composition of Lebesgue measurable function $f$, with a continuous function $g$ having a certain property, is Lebesgue measurable

Composition of 2 Lebesgue measurable functions is not lebesgue measurable: Are these two functions Borel Measurable?

Compositions preserving measurability

Remark(5). I note that, if $g:{\bf R}\rightarrow {\bf R}$ is a continuous function, then we have the following conclusion(it is a sufficiency condition): If $u: {\bf R}\rightarrow {\bf R}$ is Lebesgue measurable, then $g\circ u$ is also a Lebesgue measurable function on ${\bf R}$. This is easily proved from the identity $(g\circ u)^{\leftarrow}(U)=u^{\leftarrow}(g^{\leftarrow}(U))$, because, by continuity of $g$, we have that $g^{\leftarrow}(U)$ is an open set, and by measurability of $u$, we have that $u^{\leftarrow}(g^{\leftarrow}(U))$ is a measurable set, which proves the assertion. But I do not know if this sufficient condition is also necessary.

Remark(6). Is this the answer to the question, I am not absolutely sure:

Theorem on Measurability-Preserving Maps?

The issue here, is I believe, the following: in PhoemueX's answer to the question above, in terms of our terminology, it is shown that (B) property is exactly the property that $g$ is a Borel function (if I understood the argument correctly). So now, the problem with PhoemueX's argument in the case of (L) property is that, if we in particular consider invertible measurable functions $u:{\bf R}\rightarrow {\bf R}$ such that $u$ satisfies the Luzin (N)) property on ${\bf R}$, we can write $g=g\circ u\circ u^{-1}$, and the situation is the following: $$ g^{\leftarrow}(U)=u((g\circ u)^{\leftarrow}(U)) $$ whereby we know that $(g\circ u)^{\leftarrow}(U)$ is a measurable set. So, since we know that $u$ satisfies the Luzin (N) property, it follows that $u((g\circ u)^{\leftarrow}(U))$ is a measurable set (that the Luzin (N) property of $\psi$ is equivalent to preserving measurability of sets under the action of $\psi$ is already established here:

A measurable function $F$ sends measurable sets to measurable sets $\iff$ it has the Luzin n property. (In $R$)

In particular, by

Measurability of the inverse of a measurable function

it follows that $u^{-1}:{\bf R}\rightarrow {\bf R}$ is also a measurable function). My point is that, even if we have measurability of $u((g\circ u)^{\leftarrow}(U))$, can we conclude that $g^{\leftarrow}(U)$ is a Borel set? If we can, we completed the proof of necessity. I am not sure if this line of reasoning is correct. Maybe it is simply the case that we have to consider the smaller class of functions $u$ in order to get that $g$ is Borel function, namely the class of measurable functions $u:{\bf R}\rightarrow {\bf R}$ such that $u$ is invertible, such that $u$ satisfies the Luzin (N) property, and such that $u^{-1}$ is a Borel function WHICH SATISFIES THAT FOR EVERY MEASURABLE SET $M$ WE HAVE THAT $u(M)$ IS A BOREL SET (IF SUCH FUNCTIONS EXIST...). Then, we should get that $g^{\leftarrow}(U)$ is a Borel set. I would like to hear another opinion on this, just to be safe.

Answer 1

Pretty sure the answer you linked to in Remark (6) was talking about property (B) in your terminology. It is not hard to show property (B) is just equivalent to being Borel (to show necessity, just note the identity function $x: \mathbb{R} \to \mathbb{R}$ is Borel, and thus $g = g \circ x$ is Borel). For property (L), it is equivalent to the inverse image of any open set (or equivalently any Borel set) being universally measurable. Indeed, the sufficiency follows from observing that, if $B \subset \mathbb{R}$ is universally measurable, and $u: \mathbb{R} \to \mathbb{R}$ is Lebesgue measurable, then $u^{-1}(B)$ must be Lebesgue measurable. This is because the pushforward measure $u_\ast(\lambda)$ (where $\lambda$ is any complete probability measure defined on all Lebesgue measurable sets that is equivalent to the Lebesgue measure - say, the Gaussian measure) is a complete Borel probability measure on $\mathbb{R}$, so $B$ must be measurable w.r.t. $u_\ast(\lambda)$, which, by definition, means $u^{-1}(B)$ is Lebesgue measurable. For necessity, in order to show $B$ is universally measurable, it suffices to show that $B$ is measurable w.r.t. any atomless complete Borel probability measure $\mu$ on $\mathbb{R}$. Again, let $\lambda$ be any complete probability measure defined on all Lebesgue measurable sets that is equivalent to the Lebesgue measure. Then there is a Borel automorphism $u: \mathbb{R} \to \mathbb{R}$ s.t. $u_\ast(\lambda) = \mu$. As $u$ is Borel and thus Lebesgue measurable, $g \circ u$ is Lebesgue measurable, so we must have $u^{-1}(B)$ is Lebesgue measurable - which, again by definition, means $B$ is measurable w.r.t. $u_\ast(\lambda) = \mu$.

Remark: This implies property (L) is strictly weaker than being Borel (or equivalently property (B)), since there are analytic sets $A \subset \mathbb{R}$ which are not Borel. But analytic sets are all universally measurable, so $g = 1_A$ for an analytic but non-Borel $A$ would have property (L) but not be Borel.