Suppose that $X_n\Rightarrow X$ and $Y_n\Rightarrow Y$ as $n\to\infty$ where "$\Rightarrow$" means convergence in distribution. We know that it does NOT imply that $(X_n,Y_n)\Rightarrow (X,Y)$ as $n\to\infty$.
In the following proof, which step is incorrect?
Proof. Let $g_n(c_n)=E[f(X_n,Y_n)|Y_n=c_n]$ and $g(c)=E[f(X,Y)|Y=c]$, where $f$ is any given bounded and continuous function and $c_n\to c$ as $n\to\infty$. Because $(X_n,c_n)\Rightarrow (X,c)$ as $n\to\infty$, we have $$ g_n(c_n)\to g(c) \quad \mbox{as $n\to\infty$}. $$ Then, by continuous-mapping theorem, we have $g_n(Y_n)\to g(Y)$ as $n\to\infty$, i.e., $$ \lim_{n\to\infty} E[f(X_n,Y_n)|Y_n]=E[f(X,c)|c=Y]. $$ By dominated convergence theorem, we have $$ \lim_{n\to\infty}E[f(X_n,Y_n)]=\lim_{n\to\infty}E[E[f(X_n,Y_n)|Y_n]]=E\left[\lim_{n\to\infty}E[f(X_n,Y_n)|Y_n]\right]=E[E[f(X,c)|c=Y]]=E[f(X,Y)], $$ which shows that $(X_n,Y_n)\Rightarrow (X,Y)$ as $n\to\infty$.$~~~~~~~~~~$ QED.
Note: I use the "extended" continuous-mapping theorem in the proof where there are a sequence of functions $g_n$ instead of a function $g$, see http://www.bios.unc.edu/~kosorok/lecture09.pdf (Page 23)
This proof leads us to an incorrect conclusion. So, I am wondering which step in the proof is incorrect?
Thank you!