If all the marginal distributions are continuous, then the joint distribution is continuous? Consider a random vector $X\equiv (X_1,...,X_L)$. Assume that each $X_l$ is continuously distributed with support $\mathbb{R}$, for $l=1,...,L$. Does this imply that also $X$ should be continuously distributed (although, not necessarily with support $\mathbb{R}^L$)?
I clarify that, by support  of $X_l$, I intend the smallest closed set $\mathcal{X}$ such that $Pr(X_l\in \mathcal{X})=1$.
 A: No, if the individual random variables are continuous and thus their marginal distributions can be described using pdfs, it is not necessarily the case that they enjoy a joint pdf. A standard counterexample to the OP's "claim" is when $X \sim N(0,1)$ and $Z$ is an independent discrete random variable taking on values $\pm 1$ with equal probability. Then, $Y = ZX \sim N(0,1)$ also, but $(X,Y)$ does not have a joint pdf (measured in units of probability mass per unit area). All the probability mass lies on the lines $y=x$ and $y=-x$ and since lines have zero area, $(X,Y)$ does not enjoy a joint pdf (measured in units of probability mass per unit area).  See, for example, this answer by Macro.
A: No, a very simple counterexample is $(Z, Z)$ where $Z \sim \mathcal{Norm}(0,1)$ where the marginals are standard normal but the joint distribution is concentrated on the diagonal line $y=x$. So the joint distribution do not have a density with respect to the Lebesgue measure on the plane, but it does indeed have a density with respect to the Lebesgue density on that line $y=x$.
Another simple example, but here the marginals have a density which is positive on the segment $[-1, 1]$. Let $(X,Y)$ have the uniform distribution on the unit circle, that is, we can represent it as $X=\cos(\theta), Y=\sin(\theta)$ where $\theta \sim \mathcal{Uniform}(0, 2\pi)$. I did a simulation in R:

