Suppose that $X$ is a random variable distributed according to density $g$. and that $f$ is a separate density function. Let $U \sim Unif(0,1)$. I am wondering why:
$$ P\left(U \leq \frac{f(X)}{cg(X)}\right) = \int_{-\infty}^{\infty}\int_{0}^{f(x)/cg(x)}g(x) du dx $$
Is there a simple way to see this?
It is presumed here that $X$ and $U$ are independent r.v.'s. and also that the ratio $\frac{f(x)}{cg(x)}\leq 1$ for all $x$.
Joint density function for $(U,X)$ is the product of p.d.f.'s: $$f_{(U,X)}(u,x)=1_{(0,1)}(u)g(x).$$ Here $1_{(0,1)}(u)$ is equal to $1$ for $u\in(0,1)$ and $0$ otherwise.
Then for $D=\left\{(x,u)\,:\,u\leq \frac{f(x)}{cg(x)}\right\}$ $$ P\left(U \leq \frac{f(X)}{cg(X)}\right) = P\left((U,X)\in D \right)=\iint_D f_{(U,X)}(u,x)dudx. $$ To integrate over $D$ one can allow $x$ to run along $\mathbb R$, and for any $x$ allow $u$ to be in range $0, \frac{f(x)}{cg(x)}$. If this value is greater 1, one should integrate up to $1$.
Finally $$P\left(U \leq \frac{f(X)}{cg(X)}\right) = \int_{-\infty}^{\infty}\int_{0}^{f(x)/cg(x))}1_{(0,1)}(u)g(x) du dx= \int_{-\infty}^{\infty}\int_{0}^{\min(1,\,f(x)/cg(x))}g(x) du dx.$$
This is the same as initial integral if $\frac{f(x)}{cg(x)}\leq 1$ for all $x$.