Definition of Convergence a.s. in Counting probability

Question

If we have probability space [0,1], P(A)=|A|, we have

$$Xn=n^{1/p}*1_{[0,1/n]}$$ Show that Xn converges almost surely. (p denoting any p norm here)

So i proved Xn converges in probability to 0 since $P(|Xn|>\epsilon)<1/n$ so approches to 0 right? Then If Xn converges a.s., it has to converge to 0 a.s. However, the answer says Xn converges a.s. with no explanation; but Xn can contain infinitely many 0's and $n^{1/p}$'s. So how would it converge a.s. to 0?

Answer 1

For any $\omega$ except 0, for $n>1/\omega$ you have $X_n(\omega)=0$.