A point at which a function equals the derivative of its indefinite integral is not necessarily a Lebesgue point. Rudin, in Real and Complex Analysis, Theorem 7.11, proves that if $f\in L^1(R^1)$ and
$$F(x)=\int_{-\infty}^x f(y)\,dy$$
then $F'(x)=f(x)$ at every Lebesgue point of $f$. His definition of a Lebesgue point is a point $x$ such that
$$\lim_{r\to 0}\frac{1}{m(B_r)}\int_{B(x,r)}|f(y)-f(x)|\,dy=0.$$
However, he doesn't mention if the converse of the theorem is true or not. That is, if
$F'(x)=f(x)$ then is $x$ a Lebesgue point of $f$?
Now it turns out that nearly seven years ago, the answer was basically given as "No" in the answer to this post Derivative of the indefinite integral and Lebesgue point.
In the answer, Professor Israel writes (I'm paraphrasing and changing the function so that the lower limit of the integral defining $F$ being $-\infty$ will work):
Let
$$f(x)=\begin{cases} 2x\sin(1/x)-\cos(1/x)&\text{if $x>0$}\\
0&\text{if $x\leq 0$.}\end{cases}$$
Then
$$F(x)=\begin{cases} x^2\sin(1/x)&\text{if $x>0$}\\0&\text{if $x\leq 0,$}
\end{cases}$$
and $f(0)=F'(0)$. However, "it's not hard to show that"
$$\liminf_{\epsilon\to 0}\frac{1}{2\epsilon}\int_{-\epsilon}^\epsilon|f(y)-f(0)|\,dy>0.$$
This would indeed show that $0$ is not a Lebesgue point of $f$ and would supply the counterexample that I'm looking for. I find it certainly plausible that the "not to hard to show" statement is true, but what I'd like to ask for is a proof.
What I've got so far is that the integrand is
$\geq |\cos(1/y)|-2\epsilon$ (assuming we keep $\epsilon>0$), so what I'm looking for is an estimate for the lower bound for
$$\int_{-\epsilon}^\epsilon|\cos(1/y)|\,dy.$$
 A: I often find an example built from scratch as below easier to understand than an example that relies on the magic of sines and cosines as above. The  sort of example I have in mind can look harder at first, since you have to understand the construction instead of just looking at some algebraic formula, but once you do understand the construction it can be much simpler to see why the example does what's required.
For $n=1,2\dots$ let $$I_n=\left(\frac1{n+1},\frac1n\right].$$Define $f$  by saying $f=0$ on $\Bbb R\setminus\bigcup_nI_n$, $f=1$ on the left half of $I_n$ and $f=-1$ on the right half of $I_n$,  so in particular $$\int_{I_n}f=0.$$
If $0<\epsilon<1$ then $$\frac1{2\epsilon}\int_{-\epsilon}^\epsilon|f(t)-f(0)|\,dt=\frac12,$$so $0$ is not a Lebesgue point.
Now given $\epsilon\in(0,1)$ choose $N$ so $\epsilon\in I_N$. Then $$\frac1\epsilon\int_0^\epsilon f=\frac1\epsilon\int_{\frac1{N+1}}^\epsilon f,$$so $$\left|\frac1\epsilon\int_0^\epsilon f\right|\le\frac{|I_N|}\epsilon=\frac1\epsilon\left(\frac1N-\frac1{N+1}\right)\le (N+1)\left(\frac1N-\frac1{N+1}\right)\to0\quad(\epsilon\to0),$$since $N\to\infty$ as $\epsilon\to0$. So $F'(0)=0$.
A: Note
$$\int_0^\epsilon |\cos(1/y)|\,dy =\int_{1/\epsilon}^\infty \frac{|\cos x|}{x^2}\,dx.$$
Take $\epsilon=1/(2\pi n).$ Then the last integral is greater than
$$\sum_{k=n}^{\infty} \frac{1}{(2\pi (k+1))^2}\int_{0}^{2\pi}|\cos x|\,dx.$$
We are thus looking at a constant times $\sum_{k=n}^{\infty}\dfrac{1}{(k+1)^2}.$ The last sum is on the order of $1/n$ and this gives what you need.
