1
$\begingroup$

Let $A$ be Lebesgue measurable and $0<\lambda(A)<\infty$. Let $\alpha\in(0,1)$. Prove that there exists an open interval $P$ such that: $$\lambda(A\cap P)\leq\alpha\lambda(P)$$

I found a proof in the internet, but it is for the different inequality ($\geq$). Is this inequality correct, how should I proceed?

$\endgroup$
  • 2
    $\begingroup$ This is correct but boring. The other inequality is a special case of a more interesting fact called Lebesgue density theorem. $\endgroup$ – user203787 Jan 14 '15 at 18:26
4
$\begingroup$

If the measure of $A$ is finite, then most of the measure of $A$ must live in a sufficiently big ball. So given $\alpha$, choose a ball big enough so that the complement intersect $A$ has measure less than $\alpha$, then take $P$ to be any interval of length one outside of the ball.

$\endgroup$
  • $\begingroup$ What do you mean by saying "most of the measure must live..."? $\endgroup$ – nilcorc Jan 14 '15 at 18:36
  • $\begingroup$ As you take balls around any point and limit the radius to infinity, the measure of the ball intersect your set approaches the measure of your set. This is a consequence of dominated convergence theorem on indicator functions. $\endgroup$ – T.J. Gaffney Jan 14 '15 at 18:39
  • $\begingroup$ Can't I do it like that: I take $\alpha\in(0,1)$ and let's say $\lambda(A)=X$ and our interval $P=(0, \frac{X}{\alpha})$. So $\lambda(P)=\frac{X}{\alpha} \Rightarrow \alpha \lambda(P)=X$. $X=\lambda(A)\geq \lambda(A\cap P)$. $\endgroup$ – nilcorc Jan 14 '15 at 22:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.