Continuous mapping theorem - counterexample

Question

The continuous mapping theorem states that

Let $g: R^n \rightarrow R^k $ be continuous in every point of a set $C$ such that $\mathbb P\left(X\in C\right)=1$.

If $X_n \xrightarrow{d} X $ then ${g(X_{n})\stackrel{d}{\rightarrow}g(X).}$

But I have a counterexample for this theorem

Let $X_n$ is a sequence of two dimension random variable $X_n =(X_n^1,X_n^2)$ such as $X_n^1 = N(0,1)$ and $X_n^2 = -X_n^1$. So we have $$X_n =(X_n^1,X_n^2)\xrightarrow{d} (N(0,1),N(0,1)) $$ Given $g: R^2\rightarrow R $ and $g(x,y) = x+y$. $g$ is so a continuous function in $R^2$. Arcording to the continuous mapping theorem, we must have $${g(X_{n}) = g(X_n^1,X_n^2)\stackrel{d}{\rightarrow}g(N(0,1),N(0,1)) = N(0,1)+N(0,1) = N(0,\sqrt{2})}$$

But $$g(X_{n}) = g(X_n^1,X_n^2) = X_n^1 + X_n^2 = X_n^1 -X_n^1 = 0 $$

What is the error in my arguments?

Thank you in advance.

Answer 1

If every $X_n^1$ has standard normal distribution and $X_n^2=-X_n^1$ then: $$X_n=(X_n^1,X_n^2)\stackrel{d}{\rightarrow}(U,V)$$ where $(U,V)$ has a bivariate normal distribution such that $U$ and $V$ both have standard normal distribution and $U+V=0$.

So we have: $$g(X_n^1,X_n^2)=0=g(U,V)$$ for each $n$.