I'm working on a exercise but I'm not sure if it is correct or not.
The exercise state as follows:
Let $A \subset R$ measurable Lebesgue.
Show that at almost all the points $x \in A$ and all the collection of intervals {$I_h$} such that $x \in I_h$ for every $h$ and $\lambda(I_h)\rightarrow 0$ when $h \rightarrow 0$
$\lim_{h\rightarrow 0}\frac{\lambda(A\cap I_h)}{\lambda(I_h)}=1$
Up to now this what I've done:
if I define for all $a\leq x \leq b$ $f(x)=\lambda(A\cap[a,x]), \qquad f_n(x)=\lambda(V_n\cup[a,x])$ where I can imagine $A$ contained in $[a,b]$ without loss of generality and $V_n$ an open set conteining A such that: $\lambda(V_n)=\lambda(A)+2^{-n}$
So I have to prove that $f'(x)=1$ a.e I can show that $(f_n-f)$ is increasing, positive and $f_n-f \leq 2^{-n}$.
By applying Fubini's Theorem to the series: $\sum_n (f_n-f)$ implies that a.e $f'_n(x)-f'(x)\rightarrow 0$ but for every $x\in A,\qquad f'_n(x)=1$ for all integer $n$.
I'm not sure if this approach work properly or maybe I have to prove some results in advance.
What do you think? Does anybody has some ideas without using the Fubini's theorem?