# Limit in probability vs Almost sure limit

is it possible to create a sequence of random variables that converge to some constant in probability, but converge to infinity almost surely?

This question is motivated by the fact that for every subdivision of the interval $[0,t]$ whose mesh tends to zero, the sum of the squared increments of a Brownian motion converges to t in probability: $lim_{n \to\infty} \sum_{k=1}^{n} (B_{t_{k}^{n}}-B_{t_{k}^{n}})^{2}=t, \text{ in probability}$. However, it is possible to find a sequence of subdivisions whose mesh tends to zero, but $lim_{n \to\infty} \sum_{k=1}^{n} (B_{t_{k}^{n}}-B_{t_{k}^{n}})^{2}=\infty, a.s.$. Somehow I feel like that is a contradiction, but apparently it is not.

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If you have a sequence of partitions whose mesh tends to zero, the quadratic variation of Brownian Motion converges to $t$ in $L^2$, so a subsequence converges to $t$ a.s., so your "it is possible" sentence is wrong. – Lukas Geyer Oct 22 '12 at 1:11

No. If $X_n \to \mu$ is probabilty, then by the Borel-Cantelli lemma, it has a subsequence converging to $\mu$ a.s.$[P]$.