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

2 Answers 2

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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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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.

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