Prove that $\sin X_n$ converges in probability

Question

I need to show that knowing $X_n\rightarrow \pi$ converges in probability then $\sin X_n\rightarrow 0$ converges in probability. So I need to show that: $$\lim_{n\rightarrow \infty}P(|\sin X_n|>\varepsilon)=0$$ for each $\varepsilon>0$. How can I do that? Idea: $$\lim_{n\rightarrow \infty} P(|X_n-\pi|\ge \varepsilon)=0$$ So: $$\lim_{n\rightarrow \infty} P(|X_n-\pi|\ge \varepsilon)=\lim_{n\rightarrow \infty} P(X_n\ge\pi+ \varepsilon)+P(X_n\le\pi- \varepsilon)$$Probability is not negative so both needs to be zero. Now let's take $P(\sin X_n\ge\sin(\pi+ \varepsilon))$and $P(\sin(X_n)\le\sin(\pi- \varepsilon))$ Yet I am uncertain that I can do that. Couldn't that change inequality? $\\$I will be glad for help.

Answer 1

A good thing to know is that $X_n \to X$ in probability if and only if for every subsequence $X_{n(m)}$ there's a subsubsequence $X_{n(m_k)}$ that converges almost surely to $X$.

In your case, that means for any subsequence $X_{n(m)}$ there's a subsubsequence $X_{n(m_k)} \to \pi$ a.s. Since sine is continuous, then $\sin(X_{n(m_k)}) \to \sin(\pi) = 0$ a.s. Now, we use the other direction of the "if and only if" to conclude that $\sin(X_n) \to 0$ in probability.

Note that nothing in that argument depends on the details of your problem. In general, if $X_n \to X$ in probability and $f$ is continuous, then $f(X_n) \to f(X)$ in probability.