Is the sum of a discrete and a continuous random variable continuous or mixed? If $X$ is a discrete and $Y$ is a continuous random variable then what can we say about the distribution of $X+Y$? Is it continuous or is it mixed? 
What about the product $XY$? 
 A: Let $X$ be a discrete random variable with probability mass function $p_X : \mathcal{X} \to [0,1]$, where $\mathcal{X}$ is a discrete set (possibly countably infinite). Random variable $X$ can be thought of as a continuous random variable with the following probability density function
$$f_X (x) = \sum_{x_k \in \mathcal{X}} p_X  (x_k) \, \delta (x - x_k)$$
where $\delta$ is the Dirac delta function.
If $Y$ is a continuous random variable, then $Z := X+Y$ is a hybrid random variable. As we know the probability density functions of $X$ and $Y$, we can compute the probability density function of $Z$. Assuming that $X$ and $Y$ are independent, the probability density function of $Z$ is given by the convolution of the probability density functions $f_X$ and $f_Y$
$$f_Z (z) = \sum_{x_k \in \mathcal{X}} p_X  (x_k) \, f_Y (z - x_k)$$
A: This answer assumes that $X$ and $Y$ are independent. Here is a solution which does not need that assumption.
Edit: I am assuming that "continuous" means "having a pdf." If continuous is instead intended to mean atomless, the proof is similar; simply replace "Lebesgue null set" with "singleton set" in what follows.

Let the support of $X$ be the countable set $\{x_1,x_2,x_3\dots\}$. I will use

Lemma: A random variable $Z$ is continuous if and only $P(Z\in E)=0$ for all Borel measurable sets $E$ with Lebesgue measure zero. 

Proof: Use the Lebesgue-Radon-Nikodym theorem. $ \square$
To prove $X+Y$ is continuous, take any null set $E$, and note that
$$
P(X+Y\in E)=\sum_k P(\{Y+x_k\in E\}\cap \{X=x_k\})\le \sum_k P(Y+x_k\in E)
$$
But $Y+x_k\in E$ if and only if $Y\in E-x_k$. The shifted set $E-x_k$ is still Lebesgue null. Since $Y$ is continuous, this means $P(Y+x_k\in E)=0$, so the above summation is zero, proving $X+Y$ is continuous. 
For the question of products, the same logic applies as long as $P(X=0)=0$. If $P(X=0)=1$, then $XY$ is discrete with $P(XY=0)=1$. Otherwise, $XY$ is a nontrivial mixture. 
A: Suppose $X$ assumes values $k \in K$ with discrete distribution $(p_k)_{k \in K}$, where $K$ is a countable set, and $Y$ assumes values in $\mathbb R$ with density $f_Y$ and CDF $F_Y$.
Let $Z = X + Y$. We have
$$ \mathbb P( Z \leq z) = \mathbb P(X + Y \leq z) = \sum_{k \in K} \mathbb P(Y \leq z - X \mid X = k) \mathbb P(X = k) = \sum_{k \in K} F_Y(z-k) p_k,$$ which can be differentiated to obtain a density function for $Z$ given by 
$$ f_Z(z) = \sum_{k \in K} f_Y(z-k) p_k.$$
Now let $R = X Y$ and assume $p_0 = 0$. Then
$$ \mathbb P(R \leq r) = \mathbb P(X Y \leq r) = \sum_{k \in K} \mathbb P(Y \leq r/X) \mathbb P(X= k) = \sum_{k \in K} F_Y(r/k) p_k,$$
which again can be differentiated to obtain a density function.
However if $p_0 > 0$, then $\mathbb P(X Y = 0) \geq \mathbb P(X = 0) = p_0 > 0$, which shows that in this case $XY$ has an atom at 0.
A: Assume that $X$ takes values in a countable set $\{n_i\}_{i=1,2,\dots}$. If $Y$ is continuous, for every real number $t$
$$
{\rm P}(X+Y=t) = \sum_i{\rm P}(X=n_i,Y=t-n_i) =0,
$$
since for all $i$ we have $\{X=n_i,Y=t-n_i\}\subseteq \{Y=t-n_i\}$ and ${\rm P}(Y=t-n_i)=0$.
Therefore $X+Y$ is continuous.
If ${\rm P}(X=0)=0$ then $XY$ is continuous too, and the proof is similar:
$$
{\rm P}(XY=t) = \sum_i{\rm P}(X=n_i,Y=t/{n_i}) =0.
$$
However, in general the product $XY$ can be discrete (for instance if $X=0$), continuous (as we have seen) or mixed (take $Y$ with uniform distribution in $(0,1)$, and let $X=0$ if $Y\le 1/2$, $X=1$ if $Y>1/2$).
