Formula of conditional probability when we have discrete and continuous random variables Consider two random variables $Y,V$ where $Y$ is discrete and $V$ is continuous. Is it true that
$$
\Pr(Y=y)=\int \Pr(Y=y|V=v) f(v) dv,
$$
where  $f$ is the probability density function of $V$? I am confused because I'm mixing probability masses with densities.
Note: Imagine that I don't have the joint distribution of $(Y,V)$. I can only use the distribution of $Y$ conditional on $V$ and the distribution of $V$.

Let $(\Omega,\mathcal F,P)$ be a probability space and $V$ a continuous random variable having density $f$ defined on the space and $A$ some element of $\mathcal F$.
Then actually the question is: what is the function prescribed by:$$v\mapsto P(A|V=v)$$?????...
Because events like $\{V=v\}$ have probability $0$ we cannot use the usual definition stating that $PA\mid 
 C)=P(A\cap C)/P(C)$.
Further we want this to be a function that satisfies:$$P(A)=\int P(A\mid V=v)f(v)dv\tag1$$
Here the Radon-Nikodym derivative comes in.
Let $\nu$ denote the measure on $\sigma(V)$ prescribed by:$$B\mapsto P(A\cap B)$$Then - if $Q$ is the restriction of $P$ on $\sigma(V)$ - measure $\nu$ is absolutely continuous with respect to $Q$ (i.e. $Q(B)=0$ implies that $\nu(B)=0$).
Then according to the theorem of Radon-Nikodym a $\sigma(V)$-measurable random variable $Z$ exists with:$$\nu(B)=\int_BZdQ=\mathbb E1_BZ\text{ for every }B\in\sigma(V)$$The fact that $Z$ is $\sigma(V)$-measurable implies that a Borel measurable function $g$ exists such that $Z=g(V)$ which makes us - in special case $B=\Omega$ - arrive at:$$P(A)=\nu(\Omega)=\mathbb Eg(V)=\int g(v)f(v)dv$$So this $g$ is a suitable candidate for satisfaction of $(1)$ and we take the liberty to define it is as $P(A|V=v)$.
Well defined?... Well, not exactly because in general there are more candidates and we just picked one out. Fortunately though it can be proved that two candidates at most differ on a $\sigma(V)$-measurable $P$-null set.
This of course works for every event $A$ so also for events $\{Y=y\}$ where $Y$ is some other random variable on the space. Doing so we find:$$P(Y=y)=\int P(Y=y\mid V=v)f(v)dv$$
