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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?

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What is the set $E$? – copper.hat Dec 5 '12 at 17:02
Sorry,just a mistake of spelling – Laura Dec 5 '12 at 17:19

This follows directly from Lebesgue's differentiation theorem. Let $A \subset \mathbb{R}$ be any measurable subset. Let $f = \chi_A$. Then $f$ is a locally integrable function on $\mathbb{R}$, and by the Lebesgue differentiation theorem we have that $\lim_{\lambda(I) \to 0} \frac{1}{\lambda(I)}\int_I f(y) dy = f(x)$ for ae x, where the limit is taken over all intervals containing $x$. Evaluate the lhs.

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My exercise seems a weaker version of Lebesgue density Theorem with I_h interval instead of classic ones.Do you think I can prove it in the same way? Anyway following your hint I can write my problem as that $\lim_{\lambda(I_h)\rightarrow 0} \frac{1}{\lambda(I_h)} \int_I \chi_A dy=\chi_A(x)$ so for every $ x \in A$ it takes value 1 a.e. – Laura Dec 13 '12 at 12:50

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