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Let $A$ be a Borel set in $(\mathbb{R},\mathcal{B})$ with positive Lebesgue-measure $\lambda(A)>0$. Show that for any $\varepsilon > 0$ there is an open interval I so that $$ \lambda(A\cap I)\geq (1-\varepsilon)\lambda(I). $$

Hello!

As a Borel-set, $A$ is Lebesgue-measurable. For any Lebesgue-measurable set, there is an open set $I\supset A$ with $\mu(I\setminus A)<\varepsilon$.

Form this I follow that

$\lambda(I\cap A)=\lambda(I)-\lambda(I\setminus A)>\lambda(I)-\varepsilon$.

... I do not see how I can prove the wanted result.

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from Lebesgue density theorem you get that almost all points of $A$ have density $1$, since $\mu(A) > 0$ there exists a point $x$ for which the denisty is actually $1$, now just take a sufficiently small ball (in this case interval) around $x$ to get what you want from the definition of density.

recall that the definition of density is $$\lim_{\epsilon \rightarrow 0} \frac{\mu(I_{\epsilon} \cap A)}{\mu(I_{\epsilon})}$$ where $I_{\epsilon} = (x-\epsilon, x+ \epsilon)$

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