Proving an inverse continuous function has distribution $F$

Question

Let $F$ be a continuous distribution function on $\mathbb{R}$, with its (generalized) inverse function $$F^{-1}(u) = \inf\{x: F(x) = u, 0 < u < 1\}$$ Prove the following statements:

a.) If $U$ is uniform random variable on $(0,1)$, then $F^{-1}(U)$ has distribution function $F$.

b.) If $X$ has distribution function $F$, then $F(X)$ is uniformly distributed on $(0,1)$.

I normally do not post a question without an attempted solution but I am a bit lost here, I don't really understand what is being asked to prove. Any suggestions or comments are greatly appreciated.

EDIT:

Attempted proof a.) Suppose $U\sim U(0,1)$, then

\begin{align*} \mathbb{P}(F^{-1}(U)\leq x) &= \mathbb{P}(U\leq F(x))\\ &= \int_{0}^{F(x)}du\\ &= F(x)\\ \end{align*} Not sure if this is correct or not but thats what I was thinking a.) wanted me to prove

Answer 1

The property in b) is called probability integral transform. The proof is relatively straightforward using the cdf method.

Let $X$ be continuous r.v. with cdf $F_{X}$, then $Y=F_{X}(X)\sim\mathcal{U}[0,1]$. In fact, $F_{Y}(y)=\mathbb{P}[F_{X}(X)\leq y]=\mathbb{P}[X\leq F_{X}^{-1}(y)]=F_{X}(F_{X}^{-1}(y))=y$.