Convergence of random variables(Proof) So, I have $X_1, X_2,...$ such that $X_n\xrightarrow{\mathbb{P}}X$, and I need to prove that $X_n+c\xrightarrow{\mathbb{P}}X+c$ and $aX_n\xrightarrow{\mathbb{P}}aX$, for any $a \in \mathbb{R}$ \ $0$, and c $\in \mathbb{R}$.
My problem is that I can\t find this $\xrightarrow{\mathbb{P}}$ anywhere in my textbook, so I'm not quite sure what I'm supposed to do. Is it a probability mapping?
 A: Yes, it is "convergence in probability"- that is, the probability the sequence converges to that limit is 1.  Of course, that depends upon a given probability distribution which the determines a "measure" on the set of events.  The set of events for which that limit is correct has measure 1 and the set of events for which that limit is not correct (the sequence does not converge or converges to some other limit) has measure 0.  (Of course, in probability measure there exist non-empty sets that have measure 0.)
A: Note that the topology of convergence in probability is also metrizable, and particular $X_n$ converges in probability to $X$ if and only if
$$
\lim_{n\to \infty}\int_\Omega \min(|X_n(\omega)-X(\omega)|,1) \,P(\mathrm{d}\omega)=0.
$$
Ps. The standard definition you may be searching for is equivalent to
$$
\lim_{n\to \infty}P(\omega \in \Omega: X_n(\omega) \to X(\omega))=1.
$$
At this point, you just have to apply the definition.
A: As for the specific problem you have:
$$\mathcal{P}(|(X_{n}+c) - (X+c)|\geq \epsilon) = \mathcal{P}(|X_{n} - X| \geq \epsilon )\to 0$$
and 
$$\mathcal{P}(|aX_{n} - aX| \geq \epsilon) = \mathcal{P}(|X_{n} - X| \geq \frac{\epsilon}{a} = \epsilon_{1})\to 0$$
