Proving the Accepted Samples from Rejection Sampling follows our Posterior Distribution

Question

I get confused how Kevin Murphy gets to line $(1)$ in Machine Learning: A Probabilistic Perspective page 818 using the indicator functions. If someone can explain this to me or give me a hint that would be greatly appreciated. Thanks!

Define the following:
Let $p(x)$ be our posterior.
Let $\tilde{p}(x)$ be our unormalized posterior.
Let $q(x)$ be our proposal s.t $Mq(x) \geq p(x)$ for some constant $M>0$.
Let X be our sampled points.
Let $S$ = $\{(x,u): u\leq\frac{\tilde{p}(x)}{Mq(x)}\}$
Let $S_{0}$ = $\{(x,u): u\leq\frac{\tilde{p}(x)}{Mq(x)}, x \leq x_{0}\}$

Proof:

\begin{align*} P(X \leq x_{0} | X \text{ accepted})&= \frac{P(X \leq x_{0 }, X\text{ accepted})}{P(X\text{ accepted})} \\ &= \frac{\int\int\mathbb{1}((x,u)\in S_{0})q(x)dudx}{\int\int\mathbb{1}((x,u)\in S)q(x)dudx} \text{ (1)} \\ &= \frac{\int_{-\infty}^{x_{0}}\tilde{p}(x)dx }{\int_{-\infty}^{\infty}\tilde{p}(x)dx} \\ &= \text{CDF of p(x)} \end{align*}

Answer 1

First, the constant $M$ should be related to the unnormalised version $\tilde p(\cdot)$ and not to the normalised version $p(\cdot)$. Second, \begin{align}\int_\mathcal S q(x)\text{d}u\text{d}x &= \int_{-\infty}^\infty\int_0^1 \overbrace{\mathbb{1}_{u\le \tilde p(x)/Mq(x)}}^\text{conditional on $x$}\text{d}u q(x)\text{d}x\\&=\int_{-\infty}^\infty\int_0^{\tilde p(x)/Mq(x)}\text{d}u q(x)\text{d}x\\&=\int_{-\infty}^\infty \frac{\tilde p(x)}{Mq(x)}q(x)\text{d}x\\&=\int_{-\infty}^\infty \frac{\tilde p(x)}{M}\text{d}x\end{align} and \begin{align}\int_{\mathcal S_0} q(x)\text{d}u\text{d}x &= \int_{-\infty}^\infty\int_0^1 \mathbb{1}_{x<x_0}\overbrace{\mathbb{1}_{u\le \tilde p(x)/Mq(x)}}^\text{conditional on $x$}\text{d}u q(x)\text{d}x\\&=\int_{-\infty}^{x_0}\int_0^{\tilde p(x)/Mq(x)}\text{d}u q(x)\text{d}x\\&=\int_{-\infty}^{x_0}\frac{\tilde p(x)}{Mq(x)}q(x)\text{d}x\\&=\int_{-\infty}^{x_0} \frac{\tilde p(x)}{M}\text{d}x\end{align}