Show that $f(\xi + o_{\Bbb P}(1))=f(\xi) + o_{\Bbb P}(1)$ 
$f:\Bbb R\to\Bbb R$ is continuous, $\xi:\Omega\to\Bbb R$ a random variable, and $o_{\Bbb P}(1)$, in analogue to Landau's little-o notation, denotes any sequences of random variables on $\Omega$ that converges to $0$ in probability (implicitly indexed by $n$). Then show that 
  $$f(\xi + o_{\Bbb P}(1))=f(\xi) + o_{\Bbb P}(1).$$

In other words, we want to show $f(\xi + o_{\Bbb P}(1))-f(\xi)\to 0$ in probability as $n\to\infty$. So I investigate the probability $P(|f(\xi + o_{\Bbb P}(1))-f(\xi)|>\epsilon)$. Now if $f$ is uniformly continuous, we are easily done since we can find a uniform $\delta$ such that $[|f(\xi + o_{\Bbb P}(1))-f(\xi)|>\epsilon]$ is included in $[|o_{\Bbb P}(1)|>\delta]$. But what if the continuity is not uniform?
 A: Fix $\eta\gt 0$ and pick some $R$ such that $\mathbb P\left\{\left|\xi\right|\gt R-1 \right\}\lt\eta$. The function $f$ is uniformly continuous on $[-R,R]$, hence there exists some $\delta\gt 0$ such that if $ x,y\in[-R,R]$ and $|x-y|\lt \delta$, then $\left|f(x)-f(y)\right|\lt \varepsilon$. Now, write 
$$\mathbb P\left\{\left|f\left(X+o_{\mathbb P(1)}\right)-f\left(X\right)\right|        \gt \varepsilon \right\}\leqslant \mathbb P\left\{\left|X+o_{\mathbb P(1)}\right|\gt R\right\} +\mathbb P\left\{\left|X\right|\gt R\right\} \\+ 
\mathbb P\left(\left\{\left|X+o_{\mathbb P(1)}\right|\leqslant R\right\}\cap \left\{\left|X \right|\leqslant R\right\}\cap\left\{\left|f\left(X+o_{\mathbb P(1)}\right)-f\left(X\right)\right|        \gt \varepsilon\right\}\right)\\\leqslant  2\eta+\mathbb P\left\{\left|o_{\mathbb P(1)}\right|\gt 1\right\}+\mathbb P\left(\left\{\left|X+o_{\mathbb P(1)}\right|\leqslant R\right\}\cap \left\{\left|X \right|\leqslant R\right\}\cap \left\{\left|f\left(X+o_{\mathbb P(1)}\right)-f\left(X\right)\right|        \gt \varepsilon\right\}\cap\left\{ \left|o_{\mathbb P(1)}\right|\lt \delta\right\}\right)\\+\mathbb P\left(\left|o_{\mathbb P(1)}\right|\geqslant \delta\right).$$
But the event $$\left\{\left|X+o_{\mathbb P(1)}\right|\leqslant R\right\}\cap \left\{\left|X \right|\leqslant R\right\}\cap \left\{\left|f\left(X+o_{\mathbb P(1)}\right)-f\left(X\right)\right|        \gt \varepsilon\right\}\cap\left\{ \left|o_{\mathbb P(1)}\right|\lt \delta\right\} $$ is empty.    
A simpler proof can be given using the fact that a sequence which converges almost surely converges in probability to the same limit, and a sequence which converges in probability admits an almost surely convergent subsequence (to the same limit).
