basic Quantile proves Let this be my definition of a quantile funktion. X is a real-valued random variable. And let F be it's distribution function. then
\begin{align*}
F^{-1}(a):=\inf\{x\in \mathbb{R}: F(x) \ge a\}.
\end{align*}
Now I want to solve, how can I prove this simple equation:
\begin{align*}
F^{-1}(a)= x_0 \Leftrightarrow P(X \le x_o ) = a, \text{    almost surely}
\end{align*}

or what assumpations are needed in order to make it happen?
If $F^{-1}(a)=x_0$ and $F$ is continuous at $x_0$ then $F(x_0)=P(X \leq x_0)=a$. It can happen that $F$ is not continuous at $x_0$. In this case either $F(x_0)=a$ or else:


$\lim_{x \to x_0^-} F(x)&lt;a$
$F(x_0)&gt;a$
There is no $y$ such that $F(y)=a$.


The only way to have $F$ be discontinuous at $x_0$ is if $P(X=x_0)&gt;0$.
There is a problem in the discrete case. If two consecutive points  $x_1&lt;x_2$  are in the support of $F$ and $F(x_1)&lt;a&lt;F(x_2)$ then $F^{-1}(a)=x_2$  but $F(x_2)\neq a$. 
What the problem asks is a special  case of the first line and holds only if $a$ is in the support of $F$.