Suppose I have $\xi_1,\xi_2,\ldots,\xi_n$ real-valued continuous random variables and let $\vec{\xi} = (\xi_1,\xi_2,\ldots,\xi_n)$ a random vector, is it true that if $\xi_i$ are continuous for all $i\in\{1,\ldots,n\}$ and independent, then $\xi_1,\xi_2,\ldots,\xi_n$ are jointly continuous? That is, does the two conditions (I think in this case it would be sufficient but not necessary) imply that $\mathbb{P}_{\vec{\xi}}$ as a push-forward measure is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^n$?
The reason I ask this is because in my probability notes, we gave an example of two continuous random variables who aren't jointly continuous. Suppose $\xi\sim \mathcal{U}[0,1]$ (the uniform distribution on $[0,1]$). Let $\eta = \xi$, then $\xi,\eta$ aren't jointly continuous. The proof of this is that we consider the set $C=\{(x,x)\ |\ x\in[0,1]\}$ and consider its push-forward measure and the Lebesgue measures on $\mathbb{R}^2$. It's clear that $m_2(C)=0$ with $m_2$ as the Lebesgue measure on $\mathbb{R}^2$ as $C$ is simply a line. On the other hand, we get $$\mathbb{P}_{(\xi,\eta)}(C) = \mathbb{P}\{0\leq\xi=\eta\leq 1\}= \mathbb{P}\{\omega\in \Omega\ |\ 0\leq \xi(\omega)\leq 1\} = 1.$$ But this construction implies that the two variables are not independent because $\eta=\xi$ is given so they are the same random variable and dependent. On the other hand, if I say we have two random variables, $\xi,\eta$ such that $\xi\sim\mathcal{U}[0,a]$ and $\eta\sim\mathcal{U}[0,b]$ with $a,b>0$ and that they are independent, then I should get that $\xi,\eta$ are jointly continuous. Is that correct?
I suppose my question is that (if my above discussion is correct) is there some generalization of this to the case of all continuous independent random variables?
The question you are asking is equivalent to the following: If $P_1,P_2,\ldots,P_n$ are proability measures on $\mathbb R$ each of which is absolutely continuous w.r.t. Lebesgue measure $m$ on $\mathbb R$ is it true that the product measure $P_1\times P_2\times\cdots\times P_n$ is absolutely continuous w.r.t. Lebesgue measure $m_n$ on $\mathbb R^n$. The answer is YES. In fact if $\frac {dP_i} {dm}=f_i$ and $f(x_1,x_2,\ldots,x_n)=f_1(x_1)f_2(x_2)\cdots f(x_n)$ then $(P_1\times P_2\times\cdots\times P_n) (E)=\int_E f \, dm_n$. This can be proved easily by first showing it when $E$ is a measurable rectangle (using Fubini's Theorem) and then using standard arguments to show that it holds for all Borel sets $E$.