Cumulative distribution function of mixed variables Given the probability density function:
$
\begin{equation}
  f_{X, Y}(x, y)=\begin{cases}
    \frac{xy}{3}, & \text{if } x=1,2,3 \text{ and } 0 < y < 1.\\\\
    0, & \text{otherwise}.
  \end{cases}
\end{equation}
$
What is the c.d.f. of $(X, Y)$?
Ok, so cheking the marginals we see that X and Y are independent:
$P(X=x)=(x/3)\int_0^1 y dy \\\\ =\frac{x}{6},\:x=1, 2, 3.$
$f_Y(y)=(y/3)\sum_{x=1}^3x \\\\ = 2y,\:0 < y < 1.$
$f_{XY}(x, y)=P(X=x)f_Y(y)$
The joint cdf has a analog propertie of independence:
$F_{XY}(x, y)=F_X(x)F_Y(y)$
My question is, in this case, the required cdf is simply
$F_{XY}(a, b) = \frac{1}{3} \sum_{x=1}^{a}x\int_0^{b}y\;dy$

Yes, your CDF expression is correct for $a \in \{1,2,3\}$ and $b \in [0,1]$.
In general, for discrete $X$ and continuous $Y$ you have
$$
F_{X,Y}(a,b)= \sum_{x\leq a}\int_{-\infty}^b f_{X,Y}(x,y)dy
$$
As you pointed out, when the random variables are independent this factorises into
$$
\left(\sum_{x\leq a}f_{X}(x)\right) \left(\int_{-\infty}^b f_{Y}(y)dy\right)=F_X(a)F_Y(b)
$$
