# Lebesgue-measurable, Borel set, open set

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 $\mu(I\setminus A)<\varepsilon$.

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

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

• Did you ever get an answer for this ? Apr 22, 2021 at 14:33

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 $I_{\epsilon} = (x-\epsilon, x+ \epsilon)$

Recall that by Lebesgue's density theorem,

$$\lim_{\delta\downarrow0}\frac{\lambda(A\cap(x-\delta,x+\delta))}{\lambda((x-\delta,x+\delta))}=1$$

for $$\lambda$$-almost every $$x\in A$$. As $$A$$ has positive Lebesgue measure we can thus choose a point $$x\in A$$ so that, with $$I_\delta=(x-\delta,x+\delta)$$,

$$\lim_{\delta\downarrow0}\frac{\lambda(A\cap I_\delta)}{\lambda(I_\delta)}=1.$$

Thus, by the definition of the limit, for any $$\varepsilon>0$$ we can find a $$\delta>0$$ so that

$$\frac{\lambda(A\cap I_\delta)}{\lambda(I_\delta)}\geq 1-\varepsilon,$$

i.e.

$$\lambda(A\cap I_\delta)\geq(1-\varepsilon)\lambda(I_\delta)$$

as was desired.