Lebesgue measure in one dimension

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?

• This is correct but boring. The other inequality is a special case of a more interesting fact called Lebesgue density theorem. – 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.
• 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 $\lambda(P)=\frac{X}{\alpha} \Rightarrow \alpha \lambda(P)=X$. $X=\lambda(A)\geq \lambda(A\cap P)$. – nilcorc Jan 14 '15 at 22:19