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?

  • 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

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.

  • $\begingroup$ 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

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