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How to prove that

$$\inf\{x \in \mathbb{R}: \mathbb{P}(X \le x) > \alpha \}=\sup \{x \in \mathbb{R}: \mathbb{P}(X<x) \le \alpha \}$$

for any random variable $X$ and $\alpha \in (0,1)$?

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If $P\{X\leq x\}> \alpha$ and $P\{X<y\} \leq \alpha$ then we must have $x \geq y$ because $x < y$ would imply $\alpha <P\{X\leq x\} \leq P\{X<y\}\leq \alpha$, a contardiction. This proves that LHS $\geq $RHS. Suppose, if possible, LHS >RHS. Choose $t$ such that LHS $>t>$ RHS. Conclude that $P\{X\leq t\} \leq \alpha$ and $P\{X<t\} >\alpha$ which contradicts the fact that $P\{X\leq t\}$ is increasing.

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