TITLE: Intuitive understanding of conditional density $f_{X \mid Y}(x \mid y)$
QUESTION [5 upvotes]: Let $X$ be a continuous random variable with probability density function $f_X$. The way that I think about the meaning of $f_X$ is this: if $\Delta x$ is a small positive number then
$$
P(X \in [x,x+ \Delta x]) \approx f_X(x) \,\Delta x.
$$
Now suppose that $Y$ is also a continuous random variable. Is there a similar way to think about the conditional probability density function $f_{X \mid Y}(x \mid y)$? Notice that I cannot write
$$
\tag{1} P(X \in [x,x + \Delta x] \mid Y = y) \approx f_{X \mid Y}(x \mid y) \,  \Delta x
$$
because the event $Y = y$ has probability $0$, and I can't condition on an event with probability $0$.
Question: Since equation (1) does not make sense, how should I think about $f_{X \mid Y}(x \mid y)$?

By the way, Bertsekas's book Introduction to Probability defines $f_{X\mid Y}$ like this:
$$
f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}.
$$
However, I do not find this formula to be intuitive, despite the formula's superficial similarity with the formula
$$
P(A \mid B) = \frac{P(A \cap B)}{P(B)}.
$$

REPLY [4 votes]: Let $A = \{X \in [x,x+\Delta x]\}$ and $B = \{Y \in [y,y+\Delta y]\}$.  
By your reasoning
$$P(A|B) =\frac{P(A \cap B)}{P(B)} \approx \frac{f_{XY}(x,y) \Delta x \Delta y}{f_{Y}(y) \Delta y} = \frac{f_{XY}(x,y) }{f_{Y}(y)} \Delta x.$$
Now think of $P(X \in [x,x+\Delta x] | Y= y)$ (intuitively) as 
$$P(X \in [x,x+\Delta x] | Y= y)= \lim_{\Delta y \to 0}P(A|B) \approx \frac{f_{XY}(x,y) }{f_{Y}(y)} \Delta x, $$
where $\displaystyle f_{X|Y}(x,y) = \frac{f_{XY}(x,y) }{f_{Y}(y)}.$