geometric-measure-theoryreal-analysis
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.
This works, because: (1) Your definition of absolute continuity is equivalent to the other standard definition, namely, the condition that $|\mu_g|(A)=0$ implies $|\mu_f|(A)=0$ (your condition implies that $f\in BV$, so $\mu_f$ is well defined).
(2) By a sufficiently general version of the (Lebesgue) differentiation theorem, $\lim_{h\to 0} \mu_h(x,x+h)/|\mu_g|(x,x+h)$ exists for $|\mu_g|$-a.e. $x$ and if $\mu_h\ll |\mu_g|$, then the limit computes the Radon-Nikodym derivative $d\mu_h/d|\mu_g|$. Thus your quotient converges to $(d\mu_f/d|\mu_g| )/(d\mu_g/d|\mu_g|)$; the denominator takes the values $\pm 1$, so the division doesn't make any trouble.
In general, a Radon-Nikodym derivative satisfies $\int f (d\mu/d\nu)\, d\nu = \int f\, d\mu$. This gives the formula from part (2) of your question.
As a warm-up, let's do an example with one point of discontinuity.
Our function $f$ looks like this:
Here $f : \mathbb R \to \mathbb R$ is zero, except for a sequence of triangular spikes: height $1$, centered at $1/2^n, n=1,2,3,\dots$, with
width $1/4^n$, respectively.
Of course $f$ is continuous everywhere except $0$, and discontinuous at $0$. I claim every point is a Lebesgue point for $f$. By continuity, all nonzero points are Lebesgue points. What about $0$? For $r>0$,
$$ \frac{1}{\lambda(B_r(0))}\int_{B_r(0)}|f(0)-f(y)|\;dy =\frac{1}{2r}\int_0^r f(y)\;dy $$
Some simple estimates, $$ \int_0^1 f(y)\;dy = \frac12\sum_{n=1}^\infty\frac{1}{4^n} = \frac{1}{6} < \frac14 \\ \int_0^r f(y)\;dy \le \frac{1}{4}\qquad\text{for } \frac14 \le r \le\frac12 \\ \int_0^r f(y)\;dy \le \frac{1}{4^n}\qquad\text{for } \frac1{2^{n+1}} \le r \le\frac1{2^n}, n=1,2,3,\dots \\ \int_0^r f(y)\;dy \le 4r^2\qquad\text{for }0<r<\frac12 \\ \frac{1}{2r}\int_0^r f(y)\;dy \le 2r\qquad\text{for }0<r<\frac12 \\ \lim_{r\to 0^+}\frac{1}{2r}\int_0^r f(y)\;dy = 0 . $$
For the full question, I expect a counterexample will be almost the same. Let $E$ be a fat Cantor set. let $a_n$ be a sequence of points in the complement of $E$ that cluster everywhere in $E$; for example the centers of the open intervals removed to construct $E$. Then let $f$ be zero except for a sequence of triangluar spikes, height $1$, centered at $a_n$, width $b_n$, where $b_n \to 0$ very fast; in particular, $b_n \le \frac1{4^n}$ and $b_n < \frac14\operatorname{dist}(a_n,E)$. Proceed as above to show every point of $E$ is a Lebesgue point.
Best Answer
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.