Topologized by convergence in probability? What is the topology when it is refered to the topology of convergence in probability, or "the space is topologized by convergence in probability"?
Let say you have a space $(\Omega, \mathcal{F},P)$, and a sequence of random variables $X_n$ converging in probability to X. Then for any $\epsilon$, $P(\{\omega : |X_n(\omega)-X(\omega)| > \epsilon \})\rightarrow 0$.
But what does it mean that we equip the space of random variables with the topology of convergence in probability? What are the actual open sets in this topology? All I see is for instance that for any open set around $X$ then there exists an N such that for $n \ge N$ the sequence $X_n$ is in this open set. But this is just an observation, how are the open sets actually constructed, and is the topology unique?
A reference for my question is Protter: Stochastic integration and differential equations. On page 52 he writes: "We also write $\textbf{L}^0$ for the space of finite-valued random variables topologized by convergence in probability."
 A: If we define
$$ \rho(X,Y)=\mathbb{E}\Big[\frac{|X-Y|}{1+|X-Y|}\Big]$$
then one can show that $\rho$ is a metric on $L^0$ (up to almost sure equivalence), and that $X_n\to X$ with respect to $\rho$ if and only if $X_n\to X$ in probability. So this is one possible approach. 
A: Another approach is taking $d(X,Y)=E[|X-Y|\wedge 1]$, where $\wedge$ is for the min: $a\wedge b=\min\{a,b\}$. Then $d:L^0\times L^0\to\mathbb{R}$, up to almost sure equivalence, is a metric and the topology that it generates is convergence in probability. That is, if we denote by $X_n\stackrel{P_*}{\rightarrow} X$ convergence in $d$-metric, and by $X_n\stackrel{P}{\rightarrow} X$ as usual probability convergence, then $X_n\stackrel{P_*}{\rightarrow} X$ iff $X_n\stackrel{P}{\rightarrow} X$:


*

*Suppose $X_n\stackrel{P_*}{\rightarrow} X$. Let $\epsilon>0$. Since events $\{w\in\Omega\,|\,|X_n(w)-X(w)|\ge\epsilon\}$ and $\{w\in\Omega\,|\,|X_n(w)-X(w)|\wedge 1\ge\epsilon\wedge 1\}$ are equal, then, by Chevyshev, $$P(|X_n-X|\ge\epsilon)=P(|X_n-X|\wedge1\ge\epsilon\wedge1)\le\frac{E[|X_n-X|\wedge1]}{\epsilon\wedge 1}=\frac{d(X_n,X)}{\epsilon\wedge 1}.$$


Doing $n\to\infty$ we get $X_n\stackrel{P}{\rightarrow} X$.


*

*Now suppose $X_n\stackrel{P}{\rightarrow} X$. Let $\epsilon<1$. Note that, in general, we can calculate the expectation of a nonnegative random variable as $E[Y]=\int_0^\infty P(Y\ge x)dx$. Thus
$\begin{multline*}
d(X_n,X)=\int_0^\infty P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x\\
=\int_0^1 P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x+\int_1^\infty P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x\\
=\int_0^1 P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x,
\end{multline*}$


cause $P(|X_n-X|\wedge 1\ge x)=0$ for $x>1$. Therefore
\begin{eqnarray}
&&E[|X_n-X|\wedge1]\nonumber\\
&=&\int_0^\infty P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x\nonumber\\ 
&=&\int_0^1 P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x\nonumber\\
&=&\int_0^\epsilon P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x+\int_\epsilon^1 P(|X_n-X|\wedge 1\ge x)\,\mathrm{d}x\nonumber\\
&\le&\epsilon+(1-\epsilon)P(|X_n-X|\wedge 1\ge \epsilon)\nonumber\\
&\le&\epsilon+(1-\epsilon)P(|X_n-X|\ge \epsilon)\nonumber\\
\end{eqnarray}\
Now it is easy to conclude.
