Confusion: Conditioning a Discrete rv on a Continuous rv, "Sampling Importance Resampling" Background
In Introducing Monte Carlo Methods with R by Robert and Casella, in the discussion on "Sampling Importance Resampling", I'm confused by the following argument.
Suppose $f$ and $g$ are pdfs.  Sample $X_1 , X_2, \ldots, X_n \sim g$ and then define $X^* = X_i $ with probability $f(X_i)/(ng(X_i))$ (assume for now these probabilities sum to $1$, and I also remark that I assume the authors intend to say they are defining the conditional distribution $X^*|X_1, \ldots, X_n$ even though not stated explicitly...)
The argument is that
$
\begin{aligned}
Pr(X^* \in A)&= \sum_{i=1}^n Pr(X^* \in A \text{   and } X^* = X_i) \\
 &= \int_{A}\frac{f(x)}{g(x)}g(x) dx \\
&= \int_A f(x) dx
\end{aligned}
$
so that creating $X^*$ in this way is a valid way to create draws from $f$.
(They then talk about how the ratios $f(X_i)/(ng(X_i))$ must be normalized, but that's not important for my source of confusion.)
My Question
I'm confused by the step that $Pr(X^* \in A \text{   and } X^* = X_i) = \int_A \frac{f(x)}{n g(x)}g(x) dx = \int_A \frac{f(x)}{n} dx$ because $X_i$ is continuous and $X^*$ is discrete.  If they were both discrete, and we momentarily considered $g$ a probability MASS function,
$
\begin{aligned}
 Pr(X^* \in A \text{   and } X^* = X_i) &= Pr(X_i \in A \text{   and } X^* = X_i) \quad \text{ (The event sets are the same )}\\
&= \sum_{a\in A} Pr(X_i =a \text{   and } X^* = X_i) \\
 &= \sum_{a\in A} Pr(X_i = a)P(X^* = X_i \mid X_i = a) \\
&= \sum_{a\in A} g(a)\frac{f(a)}{ng(a)}\\
&= \sum_{a\in A} \frac{f(a)}{n}.\\
\end{aligned}
$
Now if I converted the sum to an integral, then my proof would be complete.  Thus, on a heuristic level, I know how to complete this argument.  The problem is I fundamentally don't understand how this argument translates to a continuous $X_i$.  In such a case, there is no notion of $P(X^* = X_i \mid X_i = a)$ because when $X_i$ is continuous, this leads to a division by zero $(P(X_i = a) =0)$. 
How do I understand this with continuous $X_i$?
 A: $X^*$ is not discrete; it's continuous. The $X_i$'s are continuous, too. It is $X^*|X_{1},\ldots,X_n$ that's discrete. I'll copy and paste your $\TeX$ and and write over the stuff that's incorrect. I basically just switched all your sums to integrals.
\begin{aligned}
 Pr(X^* \in A \text{   and } X^* = X_i) &= Pr(X_i \in A \text{   and } X^* = X_i) \quad \text{ (yes )}\\
&= \int_{a \in A} Pr(X_i =a \text{   and } X^* = X_i)da\quad \text{ ($X_i$ is cts )} \\
 &= \int_{a\in A} Pr(X_i = a)P(X^* = X_i \mid X_i = a)da \\
&= \int_{a\in A} g(a)\frac{f(a)}{ng(a)}da\\
&= \int_{a\in A} \frac{f(a)}{n}da.\\
\end{aligned}
Notes:


*

*the second and third lines have (joint and marginal) densities written in very terrible ways. This is on purpose. I'm doing this to make it look like your original work

*$P(X^* = X_i \mid X_i = a)$ is a pmf. The notation is fine here.

*$Pr(X_i =a \text{   and } X^* = X_i)$ is probably not a joint density. I am thinking of this like I am integrating a surface over a diagonal sliver region. Usually a jointly continuous random vector would yield probability zero if I was integrating over a region like this. It's probably not a smooth surface, though. There is mass on these diagonal slivers. 

*$Pr(X_i = a)$ is a marginal density. It should probably be written like $f_{X_i}(a)$.

*We can sum this thing together $n$ times (over $i$) because the events are disjoint
I myself would be interested in hearing about (3).
