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 $V$.

Suppose that $Y$ and $V$ are defined on a common probability space $(\Omega, \mathscr{F}, P)$, and $V$ has density $f$ with respect to Lebesgue measure,  then your conjecture is true. In fact, whether $Y$ is discrete or not is unessential for the proof, as the formula you listed is a special case of a more general relation $(3)$ to be shown below.
For any $A \in \mathscr{F}$, by the definition of conditional probability, it holds that
\begin{align}
P(A) = \int_\Omega P(A | V)dP, \tag{1}
\end{align}
where "$P(A|V)$" is a shorthand for the standard sub-$\sigma$-field-based conditional probability notation $P(A|\sigma(V))$.  Also as a part of the definition of conditional probability, $P(A|V)$ is $\sigma(V)$-measurable, implying that there exists a Borel function $g: \mathbb{R} \to \mathbb{R}$ such that $P(A|V) = g(V)$. The change of variable theorem then gives
\begin{align}
\int_\Omega P(A|V)dP = \int_\Omega g(V(\omega))P(d\omega) = 
\int_\mathbb{R}g(v)f(v)dv.  \tag{2}
\end{align}
By convention, the function $g(v)$ is usually denoted by $P(A|V = v)$.  Technically speaking, the notation makes sense because for any $v \in \mathbb{R}$, the random variable $P(A | V)$ is almost surely a constant on the set $\{\omega: V(\omega) = v\}$.  In other words, on this $\sigma(V)$-set, $P(A|V)_\omega$ almost surely equals to a constant that depends only on $v$, which is therefore reasonable to be denoted by $g(v)$ (for more discussions on how to justify the "$P(A|V = v)$" notation rigorously under the measure-theoretic probability, refer to Exercises 33.4, 33.5, 33.6 in Probability and Measure by Patrick Billingsley).
Combining $(1)$ and $(2)$ gives
\begin{align}
P(A) = \int_\mathbb{R}P(A|V = v)f(v)dv.  \tag{3}
\end{align}
Now take $A = \{Y = y\} = \{\omega: Y(\omega) = y\} \in \mathscr{F}$ in $(3)$ finishes the proof of your conjecture.
