Suppose $X$ has a continuous, strictly increasing cdf $F$. Let $Y = F(X)$. What is the density of $Y$? Then let $U \sim unif(0,1)$ and let $X = F^{-1}(U)$. Show that $X\sim F$.
The first part seems like the density of $Y$ is just the pdf $f(X)$. The wording of the problem is confusing me though.
Since $F$ is continuous and monotone increasing, it has an inverse function $F^{-1}$, i.e. for any $t\in\mathbb R$, there is a unique number $F^{-1}(t)$ such that $F^{-1}(F(t)) = t$. Recall that the inverse of a monotone increasing function is also monotone increasing. We have $$\mathbb P(Y\leqslant t) = \mathbb P(F(X)\leqslant t). $$ Since $F$ takes values in $[0,1]$, it is clear that $\mathbb P(Y\leqslant t)=0$ for $t\leqslant 0$ and $\mathbb P(Y\leqslant t)=1$ for $t\geqslant 1$. For $t\in(0,1)$ we have $$\mathbb P(F(X)\leqslant t)=\mathbb P(F^{-1}(F(X))\leqslant F^{-1}(t)) = \mathbb P(X\leqslant F^{-1}(t)) = F(F^{-1}(t)) = t. $$ It follows that $$\mathbb P(Y\leqslant t)=t 1_{(0,1)}(t) + 1_{[1,\infty)}(t), $$ and therefore the density $g$ of $Y$ is the derivative of the above: $$g(y) = 1_{(0,1)}(t). $$ In other words, $Y$ has $\mathcal U(0,1)$ distribution.
For the second part, we have for any $t\in\mathbb R$ $$ \begin{align*} \mathbb P(X\leqslant t) &= \mathbb P(F^{-1}(U)\leqslant t)\\ &= \mathbb P(F(F^{-1}(U))\leqslant F(t))\\&=\mathbb P(U\leqslant F(t))\\&=\mathbb P(U\leqslant t)\\&= F(t). \end{align*} $$
