To quote from our book Monte Carlo Statistical Methods, the most straightforward way of validating Accept-Reject sampling is to see it as simulating uniformly on a set $B$ until the simulation belongs to another set $A$. If $A\subset B$, the outcome is uniformly distributed on $A$. The details proceed as follows:
There exists a fundamental (simple!) idea that underlies the Accept-Reject methodology,
and also plays a key role in the construction of the slice sampler.
If
$f$ is the density of interest, on an arbitrary space,
we can write
\begin{equation}%\label{eq:fund}
%\int_{-\infty}^x f(x^\prime) dx^\prime = \int_{-\infty}^x \int_{0}^{f(x^\prime)} du \; dx^\prime,
f(x) = \int_{0}^{f(x)} du \,.\tag{1}
\end{equation}
Thus, $f$ appears as the marginal density (in $X$) of the joint distribution,
\begin{equation}%\label{eq:fund2}
(X,U) \sim {\cal U} \{(x,u):0<u<f(x)\}\,.\tag{2}
\end{equation}
Since $U$ is not
directly related to the original problem, it is called an auxiliary
variable.
Although it seems like we have not gained much, the introduction of
the auxiliary uniform variable in (1) has brought a
considerably different perspective: Since (2) is the
joint density of $X$ and $U$, we can generate from this joint
distribution by just generating uniform random variables on the
constrained set $\{(x,u):0<u<f(x)\}$. Moreover, since the marginal
distribution of $X$ is the
original target distribution, $f$, by generating a uniform variable on
$\{(x,u):0<u<f(x)\}$, we have generated a random variable from $f$.
And this generation was produced without using $f$ other than through
the calculation of $f(x)$! The importance of this equivalence is
stressed in the following property:
Simulating $$\mathbf{ X
\sim f(x)}$$ is equivalent to simulating $$\mathbf{ (X,U) \sim {\cal U}
\{(x,u):0<u<f(x)\} \,.}$$
For example, in a one-dimensional setting, suppose that
$$
\int_a^b f(x) dx = 1
$$
and that $f$ is bounded by $m$. We can then simulate the random pair $(Y,U) \sim
{\cal U}(0<u<m)$ by simulating $Y \sim {\cal U}(a,b)$ and $U|Y=y \sim {\cal U}(0,m)$,
and take the pair only if the further constraint $0<u<f(y)$ is satisfied.
This results in the correct distribution of the accepted value of $Y$, call it $X$, because
\begin{eqnarray}%\label{eq:fun1}
P(X \le x)&=&P(Y \le x |U<f(Y)) \nonumber\\
&=& \frac{\int_a^x \int_0^{f(y)}du \; dy}{\int_a^b \int_0^{f(y)}du \; dy} = \int_a^x f(y)\; dy.
\end{eqnarray}
This amounts to saying that, if $A \subset B$ and if we generate a uniform sample on $B$,
keeping only the terms of this sample that are in $A$ will result in a uniform sample on $A$
(with a random size that is independent of the values of the sample).
