The usual definition of convergence in probability is that $\lim_n P(|X_n-X|>\epsilon)\to 0$ for all $\epsilon>0$.
Is an equivalent definition that $\lim_n P(X_n\in B) = P(X\in B)$ for all measurable $B$?
Here's a sketch of why I think this may be true:
For any closed $K$ let $K_\epsilon = \{ x:\inf_{y\in K} |x-y|\leq \epsilon \}$. Then we have $P(X_n\in K) \leq P(X\in K_\epsilon) + P(|X_n-X|>\epsilon)$ and $P(X\in K) \leq P(X_n\in K_\epsilon) + P(|X_n-X|>\epsilon)$. Letting $n$ tend to infinity, and $\epsilon\to 0$ and considering $K=\bigcap_\epsilon K_\epsilon$ and the continuity of measure, these inequalities imply $\lim_n P(X_n\in K) = P(X\in K)$.
For arbitrary $B$, note that $B$ which satisfy $\lim_n P(X_n\in B) = P(X\in B)$ are a $\lambda$-system, and closed sets are a $\pi$-system, and the Borel sets are generated by the closed sets. Therefore Dynkin's $\pi-\lambda$ lemma lets us generalize the result.
This result compares nicely with an analogous result for convergence in distribution: $X_n\to X$ in distribution iff $\lim_n P(X_n\in B) = P(X\in B)$ for all sets with $P(X\in \partial B)=0$. It shows directly how convergence in probability implies convergence in distribution, but not conversely.
Anyway, is this right? It seems like this result would be a natural thing to put next to the usual definition, but I can't find it in Kallenberg or Durrett or Billingsley or anywhere else, which makes me wonder, am I missing something?
Let $X_{n}\stackrel{P}{\to}0$ and take $B=\mathbb{R}-\left\{ 0\right\} $.
It is quite well possible that$P\left(X_{n}\in B\right)=1$ for each $n$, but next to that we have $P\left(0\in B\right)=0$.
So $\lim_{n\to\infty} P(X_n\in B)=P(X\in B)$ extremely fails to be true in that situation.
To get a concrete example:
if e.g. $U$ has uniform distribution on $[-1,1]$ then you can take $X_n=\frac1nU$.
Sidenote (handsome by studying convergence in probability): $$X_{n}\stackrel{P}{\to}X\iff X_{n}-X\stackrel{P}{\to}0$$ So for a big deal the subject can be studied by looking at the cases like $X_n\stackrel{P}{\to}0$.