Question on Convergence in Probability I appreciate if you could guide me on this question:
Assumptions:


*

*$X_n \rightarrow^p c$: $X_n$ convrges in probability to a constant c.

*g(.) is any function that satisfies: 
$$\text{if } a_n - c = o(1) \text{ then } g(a_n) - g(c) = o(1)$$
I need to show:
$$g(X_n) - g(c) = o_p(1)$$
Thank you for guiding me on this question.

Here is my solution so far:
Consider this theorem: $X_n \rightarrow^p X$ if and only if every subsequence $n_1, n_2, \dots \in \{1,2, 3, \dots\}$, has a sub-subsequence $m_1, m_2, \dots \in \{n_1, n_2, ...\}$ such that $X_{m_j}\rightarrow^{a.s.} X$ where a.s. stands for almost surely. 
Now, since $X_n \rightarrow^p c$, then by theorem above, there exists a subsequence $X_{m_j}$ for which $X_{m_j} \rightarrow^{a.s} c$ and therefore we can write (I'm not sure about this part):
$$ X_{m_j} - c = o(1)$$
Therefore, based on the definition of g(.), we can write:
$$ g(X_{m_j}) - g(c) = o(1)$$
And this means that we have sub-subsequence of $g(X_n)$ that converges almost surely to g(c) so then by the theorem above we can conclude:
$$ g(X_{n}) \rightarrow^p g(c)$$
and this can be written as:
$$ g(X_{n}) - g(c) = o_p(1)$$
What do you think about my solution?
 A: You know that $|X_n-c|>\epsilon$ has probability converging to 0. You want the same for $|g(X_n)-g(c)|>\delta$.
Suppose you take $b<c<d$. Choose them so that $|g(x)-g(c)|<\delta$ for $b<x<d$. Now evaluate the probability that $X_n$ is not in this range.

In response to your solution: You're right about the point which is shaky. The problem is that $X_m-c=o(1)$ doesn't mean anything, because $o(1)$ is a statement about analytic limits, but the left-hand side is a random variable. 
You don't need to do anything complicated, that was the point I was driving at. You've observed that
$$X_n \to^p c \iff X_n - c = o_p(1) \iff \forall \epsilon>0: \lim_{n\to\infty} \mathbb P(|X_n - c| > \epsilon) = 0$$
Now you choose an interval $(b,d)\ni c$ such that $|g(x) - g(c)| < \delta$ for any normal, real number $x\in(b,d)$. This is the only place you need an $o(1)$-type argument. Hence
$$\mathbb P(|g(X_n) - g(c)| > \delta) \quad<\quad \mathbb P(X_n \text{ not in } (b,d))$$
But now you know that the right-hand side (which is just a number now, not a random variable or anything) tends to 0 as $n\to\infty$, from the above statement.
Hence the LHS tends to zero. But this is the definition of $g(X_n)\to^p c$.
