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Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.

For the dyadic power variation of a continuous local martingale, write ${\Delta }_i^{n}M=M_{i{2}^{-n}}-M_{(i-1)2^{-n}}$, ${\delta }_n={\max }_i|{\Delta }_i^{n}M|$ and $Q_n={\sum }_i({\Delta }_i^{n}M{)}^2$. Uniform continuity on $[0,1]$ gives ${\delta }_n\to 0$ almost surely, while the deterministic-partition characterization of quadratic variation gives $Q_n\to ⟨M{⟩}_1$ in probability. In particular $(Q_n)$ is bounded in probability.

If $p>2$, then $Z_n\le {\delta }_n^{p-2}{Q}_n$, and a factor tending to zero in probability times a sequence bounded in probability tends to zero in probability. ThusIf $1<p<2$, the reverse comparison is $Q_n\le {\delta }_n^{2-p}{Z}_n$. The assumed finite $\mathrm{limsup}{Z}_n$ makes $Z_n$ eventually bounded almost surely, so $Q_n\to 0$ almost surely. By uniqueness of a limit in probability, $⟨M{⟩}_1=0$ almost surely. Its monotonicity gives $⟨M{⟩}_t=0$ for $t\le 1$.

To finish without assuming any integrability of $M$, apply the bound from part (i) to both $M$ and $-M$, with $t=1$, and then let $b\downarrow 0$. For each $a>0$ the probability of a positive or negative excursion beyond $a$ is zero. Taking countably many $a\downarrow 0$ proves $M$ is indistinguishable from zero on $[0,1]$. The same argument in fact works for every $0<p<2$ under the stated pathwise boundedness assumption.

By Brownian scaling,where $Z$ has the standard three-dimensional multivariate normal distribution. The right-hand side tends to zero in probability, since $Z$ has no atom at the origin. Part (d) gives almost-sure convergence to $Y$, which also implies convergence in probability to $Y$. Uniqueness of a limit in probability therefore gives $Y=0$ almost surely. Hence $|B_t|\to \infty$ almost surely, proving the transience of Brownian motion in dimension at least three in dimension three.