Convergence in probability iff convergence for every bounded continuous function I'm trying to show the following:
$X_n \overset{p}{\to}X \iff f \circ X_n \overset{p}{\to} f \circ X$ for every continuous, bounded function $f$.
I can show ($\Rightarrow$) already using the usual almost-everywhere converging subsequences approach.  The other direction is giving me difficulty.
I believe I can show it if $f$ is strictly increasing: then $f$ is a homeomorphism of $\mathbb{R}$ onto some interval $(a,b)$, thus $f^{-1}$ is continuous, so it suffices to apply the continuous function $f^{-1}$ to $f\circ X_n$.  (Though if this is incorrect please do let me know.)
For the general case, however, I'm not quite sure how to proceed.  My two questions are:
1) How to show $(\Leftarrow)$ when I can't rely on $f$ being strictly monotone?
2) What is the intuition behind needing $f$ to be bounded?  I feel like this should be obvious but I can't seem to intuit why.
Many thanks.
Disclaimer: I'm pursuing self-study in Cinlar's 'Probability and Stochastics' in my spare time, this is problem 3.14.
 A: Fix $\epsilon>0$ and $r>0$. We want to show 
$$\mathbb{P}(|X_n-X| \geq r) \leq \epsilon \qquad \text{for all $n$ sufficiently large.}$$
Set
$$\chi_R(x) := (-R) \vee x \wedge R = \begin{cases} R, & x \geq R, \\ x, & x \in [-R,R] \\ -R,, & x \leq -R \end{cases}.$$
Obviously, $\chi_R$ is bounded and continuous. Moreover, as $\chi_R(x)=x$ for all $|x| \leq R$, we have
$$\mathbb{P}(|X_n-X| \geq r) \leq \mathbb{P}(|\chi_R(X_n)-\chi_R(X)| \geq r) + \mathbb{P}(|X_n| \geq R) + \mathbb{P}(|X| \geq R).$$
Now
$$\begin{align*}\mathbb{P}(|X_n| \geq R)
&= \mathbb{P}\left(|X_n| \geq R, |X| \leq \frac{R}{2} \right) + \mathbb{P} \left(|X|> \frac{R}{2} \right) \\
&\leq \mathbb{P} \left( |\chi_R(X_n)| \geq R, |\chi_R(X)|=|X| \leq \frac{R}{2} \right) + \mathbb{P} \left(| X|> \frac{R}{2} \right) \\
&\leq \mathbb{P} \left( |\chi_R(X_n)-\chi_R(X)| \geq \frac{R}{2} \right) + \mathbb{P} \left(| X|> \frac{R}{2} \right) \end{align*}$$
Combining both estimates, we obtain
$$\begin{align*} \mathbb{P}(|X_n-X| \geq r) &\leq \mathbb{P}(|\chi_R(X_n)-\chi_R(X)| \geq r)+ \mathbb{P}\left(|\chi_R(X_n)-\chi_R(X)| \geq \frac{R}{2} \right) \\ &\quad + \mathbb{P}(|X| \geq R) + \mathbb{P} \left(| X|> \frac{R}{2} \right) \end{align*}$$
Choose $R$ sufficiently large such that
$$\mathbb{P} \left(| X|> \frac{R}{2} \right) \leq \epsilon$$
For this (fixed!) $R$, it follows from our assumption that we can choose $N \in \mathbb{N}$ such that
$$ \mathbb{P}(|\chi_R(X_n)-\chi_R(X)| \geq r)+ \mathbb{P}\left(|\chi_R(X_n)-\chi_R(X)| \geq \frac{R}{2} \right) \leq \epsilon$$
for all $n \geq N$. Adding all up, we get
$$\mathbb{P}(|X_n-X| \geq r)  \leq 3 \epsilon$$
for all $n \geq N$.
Concerning boundedness: The boundedness of $f$ is not needed. Obviously, the equivalence holds also if we drop the assumption "bounded" (however, then "$\Leftarrow$" is obvious.)
