Different ways of calculating the conditional probability in the continuous case For simplicity, assume a joint pdf of 2 variables $f(x,y)$.
Say we have two events $A$ and $B$. How would one calculate:
$$\Pr[A \mid B]$$
if we have continuous pdfs in question?
I would have thought that one way to do it is as follows:
$$ \Pr[A \mid B] = \frac{\Pr[A \cap B]}{\Pr[B]} = \frac{ \int_{(x,y) \text{ s.t. } A \cap B} f(x,y) \, dA }{\int_{(x,y) \text{ s.t. } B} f(x,y) \, dA}$$
however, I thought of a different way of doing it too and was not sure if they were equivalent (and if they were, how to prove their equivalence). 
First find the conditional pdf $f(x,y \mid B)$ and then
$$ \Pr[A \mid B] = \int_{(x,y) \text{ s.t. }A \cap B} f(x,y \mid B) = \int_{(x,y) \text{ s.t. }A} f(x,y \mid B)$$
Intuitively, they seemed nearly the same, though not sure how to make this argument precise.
I was thinking that if $f(x,y \mid B)$ were an actual probability instead of just a pdf, then it seems that its clear that  just $ \Pr[A \mid B] = \sum_{(x,y) \text{ s.t. } A \cap B} p[x,y \mid B]$ however, even though it seems "analogous", was not convinced that was a sufficient argument to "prove" their equivalence. Is there a better explanation for their equivalence or not?

On a very related note, is way to calculate $f(x,y | B)$ by:
$$ f(x,y | B) = \frac{  f(x,y) }{\int_{(x,y) \ s.t. \ B} f(x,y) dA}$$
which would establish the equality I wanted (though I came up with that just by analogy, not by rigorousness or mathematics) ...  If that is correct, why is that the answer?
 A: Your intuition is on the right track. For continuous random variables $X$ and $Y$ with joint pdf $f_{X,Y}$, lets require the event $B$ to not be a null-set ( i.e. $P(B)\ne 0$ ). Then, by definition, we have for events $A$ and $B$,
$$
P\left(\,\left(X,Y\right)\in A\,\Big\vert\,\left(X,Y\right)\in B\right){}:={}\dfrac{P(A\,\cap\,B)}{P(B)}{}={}\dfrac{\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}{\bf 1}_{A\cap B}\,f_{X,Y}\,\mathrm dx\mathrm dy}{\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}{\bf 1}_{B}\,f_{X,Y}\,\mathrm dx\mathrm dy}{}={}\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}{\bf 1}_{A}\,\left(\frac{{\bf 1}_{B}\,f_{X,Y}}{\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}{\bf 1}_{B}\,f_{X,Y}\,\mathrm dx\mathrm dy}\right)\mathrm dx\mathrm dy,
$$
where you have (or want to?) defined

$$
f_{X,Y}(x,y\,\vert B){}:={}\dfrac{{\bf 1}_{B}\,f_{X,Y}}{\displaystyle\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}{\bf 1}_{B}\,f_{X,Y}\,\mathrm dx\mathrm dy}\,.
$$

By inheriting all of its properties from $f_{X,Y}$, one can check that this function is a bona-fide Radon-Nikodym derivative; it is a pdf (with respect to the Lebesgue measure) for the probability measure $P(\,\dot\,\vert\,\left(X,Y\right)\in B)$. For instance, replacing the event $A$, above, with $\mathbb{R}^2$ (the entire sample space) shows that this function integrates to $1$ over $\mathbb{R}^2$ and, therefore, over $B$. 
In passing, I also note that as a Radon-Nikodym derivative, this function is uniquely defined up to events of Lebesgue measure zero, so changing the value of this function over such events does not change the probabilities evaluated using this function.   
