If $P|x|^2 = \int |X|^2dP < \infty$ then there exists $M$ so that $P |x|^2 < (\frac{M}{2})^2 P[-\frac{1}{2}M, \frac{1}{2}M]$?

Question

Let $P$ be a probability measure in the real line such that $P|x|^2 = \int |X|^2 dP < \infty$. Then how is it possible to choose $M$ so that $P |x|^2 < (\frac{M}{2})^2 P[-\frac{1}{2}M, \frac{1}{2}M]$?

This result is used in a proof of the consistency of the $2-$means. The only assumption I see here is that $P|x|^2<\infty$, but I don't see how this implies the existence of such a $M$. I would greatly appreciate any help.

Answer 1

Note that $$\Big(\frac{M}{2}\Big)^2P\Big(\Big[-\frac{M}{2},\frac{M}{2}\Big]\Big)\to\infty$$ as $M\to\infty$, and so for any positive real number $a$ there is some $M$ such that this expression is $>a$. Now take $a=\int x^2\;dP$.