# Proof $existsalpha$ s.t. $P(X>alpha)>0$ if $P(X>0)>0$

Mathematics Asked on January 1, 2022

For probability triple $$(mathbb{R}, mathcal{B}(mathbb{R}), mu)$$ prove that for a random variable $$X$$, if $$mu(X>0)>0$$, there must be $$alpha>0$$ s.t. $$mu(X>alpha)>0$$.

So if $$X$$ is a random variable with that property, it means that $$exists$$ event $$A in mathcal{F}$$ and interval $$B=(0, infty)$$ s.t.
$$A:{omega in mathbb{R}: X^{-1}(B)= A }, mu(A)=m>0$$
Since $$A$$ is an interval, we can set its upper and lower bounds as $$beta_1, beta_2$$. Since $$A in mathcal {F}$$, we can certainly find number $$alpha^{-1}$$ such that there exist two disjoint intervals $$A_1 cup A_2=A$$ with the same measure:
$$A=A_1 cup A_2, A_1 = [beta_1, alpha^{-1}], A_2 = [alpha^{-1}, beta_2], mu(A_1)=mu(A_2)=frac{m}{2}$$
Obvisouly $$alpha^{-1} in A$$, and, since $$A$$ is a preimage of $$B, alpha^{-1} = X^{-1}({alpha})$$, and $$alpha in B$$. Therefore
$$A_2 = {omega:X^{-1}(alpha, infty)}$$
and $$mu(X>alpha) = mu(A_2)=frac{m}{2}>0$$.

I think this is correct, but the hint for the problem is to use the continuity of probabilities, which I didn’t.

The set $$E_n={X>frac1n}$$ increases to $$E={X>0}$$. By monotone convergence $$0

From this, it follows that $$mu(E_n)>0$$ for all sufficiently large $$n$$.

Answered by Jean L. on January 1, 2022

Consider the expanding sequence of sets $$A_n = { X > frac{1}{k}}$$. Notice $$bigcup_{k=1}^infty { X > frac{1}{k}} = {X > 0}$$. Since $$A_n$$ is an expanding sequence of sets, then by the Monotone Convergence Theorem for sets $$lim_{ntoinfty} mu(A_n) = muBig(bigcup_{k=1}^infty A_kBig) = mu({X > 0}) > 0$$ We then get that $$0 < muBig(bigcup_{k=1}^infty A_kBig)leq sum_{k=1}^infty mu(A_k)$$ Since the series $$sum mu(A_k) > 0$$ and each term is nonnegative, then at least one term must be positive, for otherwise $$sum mu(A_k) = 0$$.

Answered by Andrew Shedlock on January 1, 2022