Convergence of marginal distribtution Here I have a question which looks a little bit weird:
$(q_n)_n$ is sequence of probability density functions of the couple $(x,y) \in \mathbb R^2$, $p_n$ is the marginal density of $q_n$, i.e. $p_n(x) := \int q_n(x,y) dy$.
Another sequence of functions $(p_n^b)_n$ is defined by $p_n^b(x) := \int b(y) q_n (x,y) dy $, where $0 < l \leq b(y) < L, \forall y \in \mathbb R$.
Given that $(p_n)_n$ converges to $p_0$ weakly in $L^2$, what do we know about $(p_n^b)_n$? Is there any chance that $(p_n^b)_n$ also converges? It would be fine if some additional conditions need to be added.
Thanks so much. 
 A: This isn't a full answer, but it might lead you down a reasonable path to explore your question. 
First: Note that if $\lambda \in \Bbb{R}$ is any scalar, then $p_n^{\lambda} = \int \lambda q_n(x,y)\,dy = \lambda \int q_n(x,y)\,dy = \lambda p_n$ will clearly converge weakly to $p_0^{\lambda} = \lambda p_0$ by the linearity of the integral. From this we can see that the requirement of the lower bound $0 < l \leq b(y) < L$ is really not giving you any better restriction. Indeed, suppose you could infer something about $p^b_n$ for measurable $b$ which is bounded as $l \leq b(y) \leq L$ (for any arbitrary choices of $0 < l \leq L < \infty$). Suppose that $\tilde{b}$ is any bounded measurable function, say $|\tilde{b}| \leq R$, then set $b(y) =  \tilde{b}(y) + R + 1$ and you have $1 \leq b(y) \leq 2R+1$; however, by setting $\lambda = -(R+1)$, you have that $p^{\tilde{b}}_n = p^{b}_n + p^{\lambda}_n = p^{b}_n + \lambda p_n$ (again by the linearity of the integral), and so you should be able to infer the same convergence for $\tilde{b}$ as you were able to for $b$. In other words, it seems to me that whatever you were able to infer about the convergence of $p^b_n$ for $0 < l \leq b \leq L$, you should be able to infer about $p^{\tilde{b}}_n$ for any bounded measurable $\tilde{b}$.
Second: From here let's say that you can say that $p^b_n$ converges weakly in $L^2$ to $p^b_0$ for any bounded measurable function $b$. Unwinding the definitions, what this implies is the following: Given any bounded measurable functions $b, f$, 
$$
\iint f(x) b(y)  q_n(x,y) dydx \longrightarrow \int f(x) p_0^b(x) dx.
$$
Assuming that the measures $dx$ and $dy$ are sufficiently nice enough to apply Fubini-Tonelli arguments, let's define $k_n^f(y) = \int f(x) q_n(x,y) \,dx$, we then have that $k_n^f(y) \,dy$ will converge weakly (on $L^2$) to the (finite signed) measure $k_0^f(dy)$ defined by
$$
k_0^f(A) = \int f(x) p_0^{1_A}(x) \, dx
$$
where $1_A$ is the characteristic function of the measurable set $A$.
Taking this second point a bit further, this should then imply that $q_n(x,y)\,dxdy$ converges weakly to a measure $q_0(dx,dy)$ defined by
$$
q_0(A \times B) = \int 1_A(x)p_0^{1_B}(x)\,dx = \int 1_B(y) k_0^{1_A}(dy)
$$
for measurable rectangles $A \times B$.
Granted I haven't been completely rigorous and checked everything, it seems from the above reasoning (assuming it is correct) that if you can infer the convergence of $p^b_n$ for arbitrary bounded measurable function $b$, then you will be facing a situation where your original measures $q_n(x,y)\,dxdy$ needed to be weakly convergent to start with. 
