Continuous mapping theorem to show $g(x_n)$ to $g(x)$ not converges.

Question

Let $X_n$ be a random variable sequence, such that $P(X_n=1)=1/n$ and $P(X_n=1/n)=1-(1/n)$.

Let g be a function, such that

$ g(x)= 0$ if $ x\le0$, and 1 if $x>0$

Show that $g(X_n)$ not converges to $g(X)$.

Not sure how to approach this. My impression is to use the Continuous Mapping Theorem that states that continuous functions are limit preserving. Thus if $X_n \rightarrow X$, then $g(X_n) \rightarrow g(X)$

By finding the value $E(X)$ and Variance $\operatorname{Var}(X)$ and applying the Chebychev inequality, I can show that $ \operatorname{plim} X_n \rightarrow 0$ as $n \rightarrow \infty$.

But this doesn't really assist in saying anything about the convergence of $g(.)$. Something to do with $g(x)$, as a pdf does not produce a CDF? But i really do not know what i am doing.

Which conditions are required for the continuous mapping theorem to hold in this case, and how to show it?

Tips and tricks are sought to better my understanding. Thank you in advance.

Answer 1

Note that $X_n\gt 0$ surely for each $n$, hence by definition of $g$, we have $g\left(X_n\right)=1$. The random variable $X$ is not specified but we can imagine that this is the limit in probability of the sequence $\left(X_n\right)_{n\geqslant 1}$. Since $X_n\to 0$ in probability and $g(0)=0$, we do not have $g(X_n)\to g(X)$ in probability (the problem comes from the fact that $g$ is not continuous).