Does weak convergence of measures preserve independence of marginals? Let $X^n = (X^n_1, \dots, X^n_d) ~ q^n$ be a $d$-dimensional random variable, where all the components are independent. That is, $X_i \perp X_j$ for $i\neq j$, and $$q^n(X) = \prod_{i=1}^d q^n_i(X^n_i).$$
If the sequence of measures $q^n$ converges weakly to some $q^*$, then are the resulting marginals also independent? I.e., is $$q^*(X)= \prod_{i=1}^d q^*_i?$$
Edit: convergence of marginals/joints
Let the state space be $\Omega^d$, and $\Omega$ be the state space of each component (i.e. $X_i \in \Omega$ for all $i$).
If $q^n \xrightarrow{w} q$, then by definition of weak convergence,
$$\int f(X) q^n(dX) \rightarrow \int f(X) q(dX) \;\; \text{ for all } f \in C_b(\Omega^d) \;\;\; \text{ as }n\rightarrow \infty.$$
Since we can take $f$ to have any support in $\Omega^d$, the above convergence of integrals holds for any combination of components; for example if $f$ has support on the first component only, then $$\int_\Omega f(X_1) q^n_1(dX_1) \rightarrow \int_\Omega f(X_1) q(dX_1).$$
I.e. the marginals converge.  This is also true for any joint, e.g. $q^n(X_1,X_2) \xrightarrow{w} q(X_1,X_2)$.
Since $q^n(X_i|X_j)q^n(X_j) = q^n(X_i,X_j)$, and we know that $q^n(X_i,X_j) \xrightarrow{w} q(X_i,X_j)$ and $q^n(X_i) \xrightarrow{w} q(X_i)$, then it seems like $q^n(X_1|X_2) \xrightarrow{w} q(X_1|X_2)$ would be true.
Here's an attempt: for all $X_2 \in \Omega$,
\begin{align}
\int q^n(X_1|X_2)f(X_1) dX_1 &= \int \frac{q^n(X_1,X_2)}{q^n(X_2)}f(X_1)dX_1 
\\
&= \frac{1}{q^n(X_2)} \int q^n(X_1,X_2)f(X_1)dX_1 
\\
&\rightarrow  \frac{1}{q^n(X_2)} \int q(X_1,X_2)f(X_1)dX_1
\\
&\stackrel{?}{\rightarrow} \frac{1}{q(X_2)} \int q(X_1,X_2)f(X_1)dX_1 
\\
&= \int q(X_1|X_2)f(X_1)dX_1 
\end{align}
I don't think "$\stackrel{?}{\rightarrow}$" is valid under weak convergence since this would require pointwise convergence. Is this where the proof breaks?
 A: Assuming that the state space $\Omega$ is a locally compact Hausdorff space and the $\left\{q_n \right\}$ are all Radon measures, then the answer is yes.
I will do the proof for $d = 2$ as the idea is the same for arbitrary $d \in \mathbb{N}$.
Since the measures are Radon, we can replace $C_b\left(\Omega^2\right)$ with $C_c\left(\Omega^2\right)$, the space of continuous real-valued functions with compact support. Let $q_1, q_2$ be the marginals of $q$ and $f(x,y) = h(x)g(y)$ where $h,g \in C_c\left(\Omega\right)$. We can rewrite $f$ as $$f(x,y) =  \left(h(x)\mathbf{1}_{\Omega}(y)\right)\left(g(y)\mathbf{1}_{\Omega}(x)\right),$$
and integrate both side with respect to $q_n$ to get
$$ \int f(x,y) q^n\left(dxdy\right) = \int h(x)g(y)q^n\left(dxdy\right)= \int h(x) q_1^n\left(dx\right)\int g(y) q_2^n\left(dy\right) \\ 
= \int h(x)\mathbf{1}_{\Omega}(y)q^n\left(dxdy\right) \int g(y)\mathbf{1}_{\Omega}(x)q^n\left(dxdy\right). $$
Passing to the limit as $n \to \infty$ we get by weak convergence,
$$ \int f(x,y)q\left(dxdy\right) = \int h(x)\mathbf{1}_{\Omega}(y)q\left(dxdy\right) \int g(y)\mathbf{1}_{\Omega}(x)q\left(dxdy\right) = \int h(x)q_1\left(dx\right)\int g(y) q_2\left(dy\right),$$
thus
$$\int f(x,y)q\left(dxdy\right) = \int f(x,y) q_1(dx)q_2(dy).$$
Since the vector space spanned by functions of the form $f(x,y) = h(x)g(x), \; h, g \in C_c\left(\Omega\right) $ is dense in $C_c \left(\Omega^2\right)$, we can conclude that
$$ \int f(x,y) q\left(dx,dy\right) = \int f(x,y) q_1(dx)q_2(dy) \quad \forall f \in C_c\left(\Omega^2\right),$$
which gives us the independence of the marginals.
