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?
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.