$Y=F(X) \implies Y\sim U(0,1)$?

Question

I have a doubt on this theorem:

The probability integral transform states that if ${\displaystyle X}$ is a continuous random variable with cumulative distribution function ${\displaystyle F_{X}}$, then the random variable ${\displaystyle Y=F_{X}(X)}$ has a uniform distribution on $[0, 1]$.

The theorem is longer, but I don't understand this first part. If I have a continuous $F(X)$ and I set $Y=F(X)$, then this implies that $Y\sim U(0,1)$? Why?

Thanks!

Answer 1

Since $X$ is a continuous random variable, for any $y$ with $0 \lt y \lt 1$ we have an $x$ with $F_X(x)=y$,

so $F_Y(y)= P(Y\le y)= P(F_X(X)\le y)=P(X\le x)=F_X(x)=y$

meaning that $Y \sim U(0,1)$