# Does the following condition suffice to imply convergence in distribution?

We know that convergence in distribution of random variables can be characterized as follows: Suppose $\{X_n\}$ and $X$ are defined on the same probability space. $X_n \stackrel{d}{\longrightarrow} X$ if and only if for any bounded and continuous function $f$, $$\mathbb{E}[f(X_n)] \rightarrow \mathbb{E}[f(X)]$$ My Question: Suppose that for any bounded continuous function $f$ and any bounded random variable $Y$, we have $$\mathbb{E}[f(X_n)Y] \rightarrow \mathbb{E}[f(X)Y]$$ Then we readily have that for any random variable $Y$, $$\mathbb{E}[f(X_n)g(Y)] \rightarrow \mathbb{E}[f(X)g(Y)]$$ holds for all bounded continuous functions $f$ and $g$. Following the result, can we conclude that for any random variable $Y$, $(X_n, Y) \stackrel{d}{\longrightarrow} (X,Y)$, that is, for any $\varphi \in C_b(\mathbb{R}^2)$, $$\mathbb{E}[\varphi(X_n, Y)] \rightarrow \mathbb{E}[\varphi(X, Y)]$$ holds?

It seems that if any $\varphi \in C_b(\mathbb{R}^2)$ can be approximated uniformly by product of two functions in $C_b(\mathbb{R})$, we can conclude so. But I am not sure if such topological argument is true.. Any hint will be greatly appreciated!