Topology induced by convergence in probability Let $(X,\mathcal{F})$ be a measurable space. Suppose $P$ and $Q$ are probability measures on $X$ such that $Q$ is absolutely continuous with respect to $P$.
Let $\tau_P$ and $\tau_Q$, respectively, be Ky Fan topology on $X$, which is topology induce by convergence in probability.
Can we compare $\tau_Q$ and $\tau_P$, i.e., which is coarser or finer?
Thanks for your help.
 A: If it's comparable and one is "coarser or finer" then the other it should be equal by changing to an equivalent measure.
But it isn't due to the fact, that convergence in probability is metrizable by the metric $d(X,Y) = E[min(1,|X-Y|)]$.
So $X_n \to X$ in probability is the same like $L^1$-convergence of $min(1,|X_n-X|)$ to zero. 
But $L^1$-convergence need not hold for an equivalent measure change so also not convergence in probability.
A: This question is a bit weird, because $\tau_Q$ and $\tau_P$ are not really naturally defined on the same set. $\tau_P$ is naturally defined on functions identified by $P$-a.e. equivalence; $\tau_Q$ is naturally defined on functions identified by $Q$-a.e. equivalence. $P$-a.e. equivalent functions are $Q$-a.e. equivalent but in general the converse is not true. 
What is true is that if $X_n \to X$ in probability under $P$ then $X_n \to X$ in probability under $Q$. This is straightforward to prove: suppose $X_n \to X$ under $P$. Let $\epsilon_X>0,\epsilon_P>0$. Take $\delta_P>0$ such that if $P(A)<\delta_P$ then $Q(A)<\epsilon_P$. Then one may take $n$ large enough that $P(|X_n-X|>\epsilon_X)<\delta_P$, in which case we have $Q(|X_n-X|>\epsilon_X)<\epsilon_P$. Since $\epsilon_X$ and $\epsilon_P$ were arbitrary, we get the result.
As for comparing the topologies themselves, it is enough to get the result by choosing a single set on which to define $\tau_P,\tau_Q$, and then to prove that both $\tau_P$ and $\tau_Q$ are sequential. 
This is where the weirdness comes in: $P$ and $Q$ are both finite, so $\tau_P$ and $\tau_Q$ are both metrizable (and hence sequential) if we identify functions which are equal $P$-a.e. and $Q$-a.e. respectively. (The metric in question can be chosen to be $d(f,g)=\| \min \{ 1,|f-g| \} \|_{L^1}$. Note that this metric makes sense even when $f,g$ are not themselves $L^1$.) 
But in general these are different sets of equivalence classes (because there may be $Q$-null sets which are not $P$-null). So it is a little bit more delicate, but still doable, to show that there is a single set that we can put both $\tau_P$ and $\tau_Q$ on such that they are both sequential.
