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.
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 $$\lim_{n\to\infty} \mu(A_n) = \mu\Big(\bigcup_{k=1}^\infty A_k\Big) = \mu(\{X > 0\}) > 0$$ We then get that $$0 < \mu\Big(\bigcup_{k=1}^\infty A_k\Big)\leq \sum_{k=1}^\infty \mu(A_k)$$ Since the series $\sum \mu(A_k) > 0$ and each term is nonnegative, then at least one term must be positive, for otherwise $\sum \mu(A_k) = 0$.