Limit of Probability and Probability of Limit Let $\{x_k\}$ and $x^*$ be a sequence and a point in $\mathbb{R}^n$, respectively. Can we conclude that
$$\lim_{k\to\infty} \mathrm{Prob}(x_k=x^*)=1$$
and
$$\mathrm{Prob}(\lim_{k\to\infty} x_k=x^*)=1$$
are equivalent or that one implies the other?
I think the first one implies the second, but not vice versa since $x^*$ might not be part of the sequence.
 A: Neither implies the other.
To see why the first does not imply the second, I'll describe a sequence of random variables defined on $\Omega = [0, 1)$. These variables will all be either $1$ or $0$ with different probabilities. I'll outline where they're $1$, and they're $0$ elsewhere.


*

*$X_1(a) = 1$ on $[0, 1/2)$

*$X_2(a) = 1$ on $[1/2, 1)$

*$X_3(a) = 1$ on $[0, 1/4)$

*$X_4(a) = 1$ on $[1/4, 1/2)$

*$X_5(a) = 1$ on $[1/2, 3/4)$

*$X_6(a) = 1$ on $[3/4, 1)$

*$X_7(a) = 1$ on $[0, 1/8)$


etc. Notice the pattern; the next few variables will be $1$ on a set of measure probability $1/8$, and that set will shift to the right until it hits $1$; then, the next few variables will be $1$ on a set of probability $1/16$, and so on.
Note that these random variables satisfy your first condition; specifically, they converge to $0$ in probability. That is,


*

*$\mathbb P(X_1 = 0) = 1/2$

*$\mathbb P(X_3 = 0) = 3/4$

*$\mathbb P(X_7 = 0) = 7/8$

*$\mathbb P(X_{15} = 0) = 15/16$


and that $\mathbb P(X_k = 0)$ is a nondecreasing sequence that tends to $1$. However, for no fixed $a \in [0, 1)$ is it the case that $X_k(a) \to 0$, because that sequence of numbers will oscillate infinitely many times between $0$ and $1$.
As you noted, the reverse implication doesn't hold either; pick a deterministic sequence that converges to something that's not in the sequence, e.g. $x_k = 1/k$ and $x^* = 0$.
A: Let $x_k=\frac 1 k$ a.s., then $P(x_k=0)=0 \to 0$ but $P( \lim_{k \to \infty} x_k=0)=1$. 
Now let $(x_k)$ being a sequence of independent r.v. being $0$ with probability $1-1/k$ and 1 with probability $1/k$. Then :
$$\lim_{k\to\infty} P(x_k=0)=\lim_{k\to\infty} 1-\frac 1 k = 1$$
But $$P(\lim_{k\to\infty} x_k=0)=P(\exists N, \forall k>N, x_k=0)$$
And
$$P(\exists N, \forall k>N, x_k=0)=P(x_k=1 \text{ finitely often })$$
But since $ \sum P(x_k=1)=\sum \frac 1 k = \infty$, by second Borel-Cantelli lemma :
$$P(\lim_{k\to\infty} x_k=0)=0.$$
