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:*


1. Here $g \circ f$ is the composite function ”f then g”.
2. We use the “strong” definition of Lebesgue points, as given [here](https://en.m.wikipedia.org/wiki/Lebesgue_point).



Does postcomposition with an absolutely continuous function preserve Lebesgue points?
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.


