probability - Special random variable on (0, 1)

Let $F$ be a distribution function and define $Y:[0, 1] \to \mathbb{R}$ as $Y(\omega) = \inf\{x: \omega \leq F(x)\} $. Show that $Y(w) \leq x$ if and only if $\omega \leq F(x)$. Deduce that $Y$ is a random variable over $((0,1),\mathcal{B}((0, 1)), Leb)$ with distribution function F.

I tried doing this by contradiction for the first part but it didn't seem to get me anywhere and im stuck. Further, does this random variable have a special name? I feel like this should be easy... Any help would be appreciated.

Answer

If $w \leq F(x)$ then it is obvious that $Y(w) \leq x$. For the converse we have to use right continuity of $F$. Let $Y(w) \leq x$. The for any $n$ there exists $y$ such that $x+\frac 1 n >y$ and $w \leq F(y)$. This gives $F(x+\frac 1 n) \geq w$. Letting $n \to \infty$ we get $F(x) \geq w$.