# A basic question concerning convergence in probability

I have a rather basic question concerning convergence in probability. It might be silly, but I'm not sure I'm getting to the correct conclusion.

Let $$X_n$$ be a sequence of real valued random variables with absolutely continuous distribution and let $$x_0$$ be a constant. Assume that $$X_n$$ converges to $$x_0$$ in probability, i.e. $$\lim_{n \to \infty}\mathbb{P}(|X_n-x_0|>\epsilon)=0, \quad \forall \epsilon>0.$$

Question: What can we say about $$\lim_{n \to \infty}\mathbb{P}(X_n? Can we claim that $$\lim_{n \to \infty}\mathbb{P}(X_n?

Of course, since $$X_n$$ also converges in distribution to a degenerate random variable whose distribution is a Dirac delta at $$x_0$$, for any $$x_{-} we would have that $$0=\lim_{n\to \infty}\mathbb{P}(X_n \leq x_{-}) \leq \liminf_{n \to \infty}\mathbb{P}(X_n But sice $$x_0$$ is not a continuity point of the limiting distribution, I'm not sure that one can conclude that the answer to the above question is "yes", just letting $$x_{-}\uparrow x_0$$. Any comments in this regard? Does the limit in my question even exist?

Let $$X$$ have standard normal distribution and $$X_n=x_0-\frac {X^{2}} n$$. Then $$X_n \to x_0$$ almost surely, hence also in probability. But $$P(X_n for all $$n$$.
By taking $$x_0-\frac {X^{2}} n$$ for $$n$$ even and $$x_0+\frac {X^{2}} n$$ for $$n$$ odd we get an example where the limit does not exist.
You can actually get $$\lim_{n\to \infty}\mathbb{P}(X_n < x_0) = \alpha$$ for any $$\alpha \in [0,1]$$. Indeed, let $$x_0 = 0$$ and $$\alpha \in [0,1]$$ and consider the random variables $$(X_n)$$ with cumulative distribution function $$F(x) = \begin{cases} 0 &, x < -1/n,\newline \alpha nx + \alpha &, - 1/ n \le x < 0, \newline (1-\alpha)nx + \alpha &, 0 \le x < 1 / n,\newline 1 &, \text{ else}. \end{cases}$$ In words, $$X_n$$ is uniformly distributed in $$[-1/n,0]$$ with probability $$\alpha$$ and in $$[0,1/n]$$ with probability $$1-\alpha$$. Then $$(X_n)$$ converges to $$0$$ in probability, but $$\mathbb{P}(X_n < 0) = \mathbb{P}(- 1/ n \le X_n < 0) = \alpha$$ for all $$n\in\mathbb{N}$$. Of course, you get the general case by considering $$Y_n = X_n + x_0$$.