Does postcomposition with an absolutely continuous function preserve Lebesgue points? Let $f: \mathbb R^n \to \mathbb R$ be a bounded measurable function, and $g: \mathbb R \to \mathbb R$ an absolutely continuous function.
Question: Is it true that if $x \in \mathbb R^n$ is a Lebesgue point of $f$, then $x$ is a Lebesgue point of $g \circ f$?
Notes:

*

*Here $g \circ f$ is the composite function ”f then g”.


*We use the “strong” definition of Lebesgue points, as given here.
 A: We only need $g$ to be continuous. For $\varepsilon>0$ we can find $u\in C^1(\mathbb{R})$ such that $$\sup_{s\in\mathbb{R}}|g(s)-u(s)|<\varepsilon .$$
For all $h>0$
$$\frac{1}{2h}\int_{x-h}^{x+h}|g(f(t))-g(f(x))|d{t}\le 2\varepsilon+\frac{K}{2h}\int_{x-h}^{x+h}|f(t)-f(x)|d{t}$$
where $K=\sup_{s\in H}|u'(s)|<\infty$ and $H$ is any bounded convex set containing $f(\mathbb{R})$. This implies if $x$ is a lebesgue point of $f$ then $$\limsup_{h\to 0 }\frac{1}{2h}\int_{x-h}^{x+h}|g(f(t))-g(f(x))|d{t}\le 2\varepsilon$$ since $\varepsilon$ was arbitrary we are done.
A: Edit: This doesn't answer the actual question, but a modified version where we don't assume that $f$ is bounded. (I overlooked this assumption originally.)
No. We can take
$$
f(x)= \begin{cases} n & a_n-d_n<x<a_n \\ 0 & \textrm{otherwise} \end{cases} ,
$$
with $a_n,d_n\to 0$ (and we of course also assume that the intervals $(a_n-d_n,a_n)$ are disjoint).
Then $x=0$ can be made a Lebesgue point of $f$ because $f(0)=0$ and
$\int_{-{a_N}}^{a_N} |f(t)|\, dt=\sum_{n\ge N} nd_n$. If $d_n\to 0$ rapidly,
then this will be $\simeq Nd_N$. If now also $a_N\to 0$ sufficiently slowly, then this will be $o(a_N)$. Of course, $\int_{-h}^h |f(t)|\, dt$ for a general $h>0$ can be estimated similarly.
(Or, to make this more concrete, we could take $d_n=2^{-n}$, $a_n=1/n$.)
However, there is no reason for $x=0$ to be a Lebesgue point of $g\circ f$ because the assumption that $g$ is absolutely continuous gives no control whatsoever on the values $g(n)$. In fact, for this same reason, $g\circ f$ need no be locally integrable near $x=0$.
