Let $Z$ be a random variable. Then the condition $$\inf\{t: P(Z \leq t)\geq 1-a\} \leq 0$$ is equivalent to $P(Z \leq 0) \geq 1- a$.
I am able to see why this is true by intuition but would love to see formal proof/justification.
Below is my attempt: Begin by noting that $\inf\{t: P(Z \leq t)\geq 1-a\} \leq 0$ is equivalent to $ P(Z \leq t) \geq 1- a $ for all $t \leq 0$. Since $\{Z \leq t\} \subseteq \{Z \leq 0\}$ for all $t\leq 0$, we have $$ P(Z \leq t) \leq P(Z\leq 0). $$ Thus, it implies that $P(Z \leq 0) \geq 1-a.$
Is my attempt above look fine? Also, I failed to cook up a proof for the other direction, any help would be appreciated.
Let $\{t:P(Z\leq t)\geq 1-a\}=S$. Note that if $t'\in S$, i.e. $P(Z\leq t')\geq 1-a$, then for all $t\geq t'$, $1-a\leq P(Z\leq t')\leq P(Z\leq t)$. Thus,
$t'\in S$ implies $t\in S$ for all $t\geq t'$ (1).
Now let $s = \inf S$. By definition, $s\leq t$ for all $t\in S$, and for all $s'>s$, there is some $t\in S$ such that $t < s'$. If $s<0$, then there is $t\in S$ such that $t<0$. By (1), this means $0\in S$. If $s=0$, we need to be very careful. It may be possible that $s\notin S$ (for example, $S= (0,\infty)$). So suppose $s\notin S$. Then $P(Z\leq 0) < 1-a$, while $P(Z\leq t)\geq 1-a$ for all $t>0$. However, the probability distribution is continuous, so $P(Z\leq 0) < 1-a$ implies that $P(Z\leq \varepsilon)<1-a$ for small $\varepsilon$, which is a contradiction.
If $P(Z\leq 0)\geq 1-a$, then $0\in S$, so $s\leq 0$.