Can we define $P(A \mid Y = y) = \lim_{\Delta y \to 0} P(A \mid Y \in [y, y + \Delta y])$? If $A$ and $B$ are events and $B$ has probability $0$, then $P(A \mid B)$ is undefined. (By the usual definition of conditional probability, we'd be dividing by zero.) However, I'm wondering if in the special case of the event $Y = y\;$ (where $Y$ is a continuous random variable), we can define
$$
\tag{1} P(A\mid Y = y) = \lim_{\Delta y \to 0} P(A \mid Y \in [y, y + \Delta y ]).
$$
(Let's assume that $Y$ has a PDF $f_Y$ that is continuous, and that $f_Y(y) > 0$.)
I think this would be nice because if $X$ is a continuous random variable then we could define a conditional PDF of $X$ given that $Y = y$ to be a PDF for $X$ with respect to the probability measure $Q(A) = P(A \mid Y = y)$ (assuming that $Q$ really is a probability measure). 
(I am currently dissatisfied with the approach to defining conditional PDFs that I read in undergrad probability textbooks, where we directly define $f_{X \mid Y}(x \mid y) = f_{X,Y}(x,y)/f_Y(y)$, because in this approach it is not clear that $f_{X \mid Y}(\cdot \mid y)$ is a PDF for $X$ with respect to some probability measure.)
Questions:


*

*Is the limit on the right in (1) guaranteed to exist? (Under mild assumptions.)

*If the limit on the right in (1) is guaranteed to exist, is the function $Q(A) = P(A \mid Y = y)$ defined as above a probability measure? (Under mild assumptions.)
What assumptions are necessary in order to have positive answers to the above questions?
 A: We define the probability measure $Y$ on $(0,1)$ as follows. Let $Y$ be uniform on $(\frac{1}{2},1)$ with total weight $\frac{1}{2}$; uniform on $(\frac{1}{3},\frac{1}{2})$ with total weight $\frac{1}{4}$; uniform on $(\frac{1}{4},\frac{1}{3})$ with total weight $\frac{1}{8}$; uniform on $(\frac{1}{5},\frac{1}{4})$ with total weight $\frac{1}{16}$; etc. Let $A$ be the event that $\lfloor \frac{1}{Y} \rfloor$ is odd. Then $\lim_{\Delta y \to 0} P(A | Y \in [0,\Delta y])$ does not exist (look at $\Delta y = 1,\frac{1}{2},\frac{1}{3},\dots$).
A: Although $(1)$ captures the essence of the conditional probability in your case, your definition $(1)$ is a bit unsatisfactory in the context of probability theory. A more systematical definition is as follows:
Let $A$ be an event and $Y$ be a random variable. Then $\mu$ and $\nu$, defined by
$$ \mu(B) = P(Y \in B), \qquad \nu(B) = P(A \cap \{Y \in B\})$$
for all Borel $B$, are Borel measures on $\mathbb{R}$ such that $\nu$ is absolutely continuous with respect to $\mu$. So there exists a non-negative function $h$ such that $ \nu(B) = \int_B h(x) \, \mu(\mathrm{d}x) $ for all Borel $B$, the fact which we often formally write as $\nu(\mathrm{d}x) = h(x) \, \mu(\mathrm{d}x)$. We denote any such function $h(\cdot)$ as $P(A\mid Y = \cdot)$. In measure-theoretic term, $h$ is called the Radon-Nikodym derivative of $\nu$ with respect to $\mu$. As the name suggests, it is indeed true that
$$ h(y) = \lim_{\delta \to 0^+} \frac{\nu([y-\delta, y+\delta])}{\mu([y-\delta, y+\delta])} = \lim_{\delta \to 0^+} P(A \mid Y \in [y-\delta, y+\delta]) $$
holds for $\mu$-almost every $y$ in the support of $\mu$, which is the statement of Lebesgue differentiation theorem applied to this particular case. This answers your question.
The second question is more delicate. The issue of the above definition is that the function $P(A \mid Y = \cdot)$ is defined only in $\mu$-a.s. sense for each given event $A$. And under such sense, it often makes less sense to discuss the value $P(A \mid Y = y)$ at a given point $y$. A similar phenomenon is that PDF is not uniquely determined. Indeed, its modification at finitely many points makes no difference in its role in probability theory, and so, the value of the PDF at a single point makes not much sense. For this reason, the above construction is incapable of defining the map $A \mapsto P(A\mid Y = y)$ as measure in general setting. This problem can be overcome by introducing the concept of regular conditional probability.
