This is a question is motivated from a problem I gave in a test.
Background: Suppose $\mu$, $\mu_n$, $n\in\mathbb{N}$, are probability measures on the real line with the Borel $\sigma$-algebra such that $\mu_n\Longrightarrow\mu$ as $n\rightarrow\infty$, i.e.,
$$\lim_\limits{n\rightarrow\infty}\int_\mathbb{R}f\,d\mu_n=\int_\mathbb{R}f\,d\mu$$
for all continuous bounded function $f$. It is well known that of $g:\mathbb{R}\rightarrow\mathbb{R}$ is continuous, then the sequence of pushforward measures $g_*\mu_n$ defined as $g_*\mu_n(A)=\mu_n(g^{-1}(A))$, converges to the pushforward measure $g_*\mu$. This follows from the fact that $f\circ g$ is bounded and continuous for any bounded and continuous $f$.
The continuity assumption on $g$ can be relaxed if the set of discontinuities of $g$ is a set of $\mu$-meaasure $0$.
I am looking for a counter-example using Bernoulli type of random variables $X_n$ in which the continuity assumption on $g$ is dropped, and $g_*\mu_n$ fails to converge to $g_*\mu$ in distribution. Here $\mu_n$ means the distribution of $X_n$ and $\mu$ is the distribution of $X$.
What I have in mind is to consider Bernoulli random variables $X_n$ such that $\mu_n(\{1/n\})=\mathbb{P}[X_n=1/n]=1/2=\mathbb{P}[X_n=1-1/n]=\mu_n(\{1-1/n\})$. This sequence converges to a the standard $0-1$ Bernoulli random variable $X$. But I have not been able to workout a suitable $g$.
A hint would suffice. Thank you!
Best Answer
The function $g=\mathbb{1}_{(0,1)}$, which takes values $0$ outside $(0,1)$ and $1$ in $(0,1)$ will work for your purposes.
Notice that the points of discontinuities of $g$ is $\{0,1\}$, and $\mathbb{P}[X\in\{0,1\}]=1$. Furthermore, $Z_n=g(X_n)=1$ and $Z=g(X)=0$.
In terms of measures (no random variables involved) define the sequence of probability measures $\mu_n$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ as $$\mu_n=\frac12\delta_{\frac1n}+\frac12\delta_{1-\frac1n}$$
For any $f\in\mathcal{C}_b(\mathbb{R})$ (here $\mathcal{C}_b(\mathbb{R})$ is the space of real valued bounded continuous functions on $\mathbb{R}$) $$\int f\,d\mu_n=\frac12 f(1/n)+\frac12 f(1-1/n)\xrightarrow{n\rightarrow\infty}\frac12(f(0)+f(1))$$ Consequently $\mu_n\stackrel{n\rightarrow\infty}{\Longrightarrow} \frac12(\delta_0+\delta_1)=:\mu$.
With $g(x)=\mathbb{1}_{(0,1)}(x)$ on $\mathbb{R}$, and $f\in\mathcal{C}_b(\mathbb{R})$ $$\int f\circ g\,d\mu_n=\frac12\big(f(g(1/n))+f(g(1-1/n))\big)=f(1),\qquad n\geq2$$ Thus, $\mu_n\circ g^{-1}\Longrightarrow\delta_1$ however, $\mu\circ g^{-1}=\delta_0$. As I mentioned before, that set the points of discontinuity of $g$ is given by $D=\{0,1\}$, and $\mu(D)=1$.