Does Weak Convergence in Joint Imply Weak Convergence in Marginal and Conditional distributions? Apologize for the confusing title.
Assume there is a sequence of two random variables $x_n,y_n$ and two random variables $x_0,y_0$ s.t.
$$(x_n,y_n)\overset{d}{\to}(x_0,y_0)$$
and that both $x_n,y_n$ and $x_0,y_0$ have continuous joint/marginal/conditional p.d.f., does the following also hold?
$$x_n\overset{d}{\to}x_0$$
and for any fixed $a\in\mathcal{Y}$, 
$$(x_n|y_n=a)\overset{d}{\to}(x_0|y_0=a)$$
If the above is not always true, then when will the above holds?
 A: Marginal yes, conditional not that I know.  Here's why for margins.  Suppose $(X_n,Y_n)$ converge in distribution to $(X,Y)$.  That means, for all continuous bounded functions $f$, we have $\lim_{n\to\infty}Ef(X_n,Y_n)=Ef(X,Y)$. Among the continuous bounded functions of 2 variables are those that depend only on their first argument: $f(x,y)=g(x), \forall y$.  Specializing to such $f$, we see that $Eg(X_n)\to Eg(X)$, so $X_n$ converges in distribution to $X$.  Alternatively, by the "continuous mapping theorem" applied to the continuous function $\pi:(x,y)\mapsto x$.  
Here is a sort of counterexample for conditional distributions.  Let $U$ and $V$ be iid uniform on $[0,1]$, write $U=\sum_{n>0} B_n 2^{-n}$ where $B_n=\lfloor U 2^n\rfloor \bmod 2$ is the $n$-th digit in the binary expansion of $U$.  Let $(X_n,Y_n) = (U, B_n + V)$.  Then it is easy to see that the distribution of $(X_n,Y_n)$ converges weakly to the uniform distribution on $[0,1]\times[0,2]$, and the marginal distributions of $X_n$ and $Y_n$ are in fact the same as the limiting margins.  But the conditional distributions are all either $\mu_0$ or $\mu_1$, where $\mu_x(A) = \lambda( A \cap [k,k+1])$ where $\lambda$ is Lebesgue measure: $P(Y_n\in A|X_n) = \mu_{B_n}(A)$.  With probability $1$ the binary expansion of $U$ has infinitely many $0$s and infinitely many $1$s in it, and thus with probability $1$ the  sequence $P(Y_n\in A|X_n)$ oscillates between two fixed conditional measures, without converging to either one.
Since  almost all $a\in[0,1]$ are such that there are infinitely many of both digits in its binary expansion, the sequence of conditional measures $A\mapsto P(Y_n\in A| X_n=a)$ oscillates the same way.
