Show Y has a uniform distribution if Y=F(X) where F(x)=P[X $\le$ x] is continuous in x.

If $F(x) = P[X\le x]$ is continuous in x, show that $Y=F(X)$ is measurable and that $Y$ has a uniform distribution $P[Y\le y] = y, \; 0\le y \le 1$

My first question is about notation. What does $F(X)$ mean? I cant make sense of $F(X) = P[X \le X]$.

Also how do you show $Y$ is measurable?

For the last part if $F^{-1}$ exists then we get that $P(Y \leq y ) = P(F(X) \leq y) = P(X \leq F^{-1}(y)) = F(F^{-1}(y)) = y$. Which would show that $Y$ has a uniform distribution on $[0,1]$. But how do I know that $F^{-1}$ exists. Wouldn't $F$ need to be strictly increasing for the inverse to exist? But we only know that it is non decreasing.

$F(x) = P(X\leq x)$ is the CDF of $X$ and is a regular old function (well, a nondecreasing right continuous function whose limit as you tend to $-\infty$ is zero and to $\infty$ is one). Borel Measurability follows from monotonicity (see this question, for example).
$F(X)$ is the random variable you get when you take the function $F(x)$ and plug $X$ into it. For example, if you have $X \sim Exp(1)$, $F(x) = \begin{cases} 1 - e^{-x} & x \geq 0 \\ 0 & o.w. \end{cases}$. Then, $F(X)$ is the random variable given by \begin{cases} 1 - e^{-X} & X \geq 0 \\ 0 & o.w. \end{cases}.
As for the second part of your question, you can define the generalized inverse $F^{-1}(y) = \inf\{ x: F(x) \geq y\}$, which is well defined and measurable for $y \in(0,1)$.
• Would this work to prove $Y$ is measurable. $X$ being a random variable is a measurable function and $X:(\Omega , \mathcal B) \rightarrow (\mathbb R , \mathcal B (\mathbb R))$ and $F$ is also a measurable function and $F:(\mathbb R , \mathcal B (\mathbb R)) \rightarrow ([0,1] , \mathcal B (\mathbb [0,1])$ so the composition $F(X)$ is measurable. – alpastor Mar 10 '16 at 5:13