You should think of the algorithm as producing draws from a random variable, to show that the algorithm works, it suffices to show that the algorithm draws from the random variable you want it to.
Let $X$ and $Y$ be scalar random variables with pdfs $f_X$ and $f_Y$ respectively, where $Y$ is something we already know how to sample from. We can also know that we can bound $f_X$ by $Mf_Y$ where $M\ge1$.
We now form a new random variable $A$ where $A | y \sim \text{Bernoulli } \left (\frac{f_X(y)}{Mf_Y(y)}\right )$, this takes the value $1$ with probability $\frac{f_X(y)}{Mf_Y(y)} $ and $0$ otherwise. This represents the algorithm 'accepting' a draw from $Y$.
Now we run the algorithm and collect all the draws from $Y$ that are accepted, lets call this random variable $Z = Y|A=1$.
To show that $Z \equiv X$, for any event $E$, we must show that $P(Z \in E) =P(X \in E)$.
So let's try that, first use Bayes' rule:
$P(Z \in E) = P(Y \in E | A =1) = \frac{P(Y \in E \& A=1)}{P(A=1)}$,
and the top part we write as
\begin{align*}P(Y \in E \& A=1) &= \int_E f_{Y, A}(y,1) \, dy \\ &= \int_E f_{A|Y}(1,y)f_Y(y) \, dy =\int_E f_Y(y) \frac{f_X(y)}{Mf_Y(y)} \, dy =\frac{P(X \in E)}{M}.\end{align*}
And then the bottom part is simply
$P(A=1) = \int_{-\infty}^{\infty}f_{Y,A}(y,1) \, dy = \frac{1}{M}$,
by the same reasoning as above, setting $E=(-\infty, +\infty)$.
And these combine to give $P(X \in E)$, which is what we wanted, $Z \equiv X$.
That is how the algorithm works, but at the end of your question you seem to be concerned about a more general idea, that is when does an empirical distribution converge to the distribution sampled from? This is a general phenomenon concerning any sampling whatsoever if I understand you correctly.
In this case, let $X_1, \dots, X_n$ be iid random variables all with distribution $\equiv X$. Then for any event $E$, $\frac{\sum_{i=1}^n1_{X_i \in E}}{n}$ has expectation $P(X \in E)$ by the linearity of expectation.
Furthermore, given suitable assumptions you could use the strong law of large numbers to show that the empirical probability converges almost surely to the true probability.
Best Answer
If you target $f$ with $g$, and you know $f(x) \le g(x)c$, then \begin{align*} P(\text{accept proposal}) &= P\left( U \le \frac{f(X)}{g(X)c} \right) \\ &= E\left( \mathbf{1}\left[U \le \frac{f(X)}{g(X)c}\right] \right) \tag{*}\\ &= E\left( E\left[ \mathbf{1}\left\{U \le \frac{f(X)}{g(X)c}\right\} \bigg\rvert X \right]\right) \\ &= E\left( P\left[ U \le \frac{f(X)}{g(X)c} \bigg\rvert X \right]\right) \\ &= E\left( \frac{f(X)}{g(X)c} \right) \\ &= c^{-1}\int f(x)g(x)/g(x) \text{d}x \\ &= c^{-1}. \end{align*}
Note that (*) is a double integral because both $U$ and $X$ are random, so you must integrate their joint density over the appropriate region. They are independent, so their joint density factors, so the joint density is $f_U(u)g_X(x) = 1 \cdot g(x) = g(x)$. There is no $u$ in this expression because it is a uniform random variable--just make sure you integrate over the right bounds.