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For $m>0$, the Brownian reflection principle givesas $t\to \infty$. Hence Brownian motion hits every positive integer almost surely. Applying the same argument to $-B$ shows that it hits every negative integer almost surely. Taking the countable intersection of these probability-one events provesArbitrarily late positive and negative values occur, and path continuity forces a zero between successive values of opposite sign. Thus recurrence of one-dimensional Brownian motion gives infinitely many visits to zero.

The pair $(B,\tilde{B})$ is a Brownian motion in ${\mathbb{R}}^{2d}$. Assume inductively that $T_{i-1}<\infty$. By the Strong Markov property at $T_{i-1}$, the differenceis a one-dimensional Brownian motion with variance rate two, started from its current value. By recurrence of one-dimensional Brownian motion, it hits zero in finite time almost surely. Hence $T_i<\infty$. Induction through $i=1,\dots ,d$ proves $T_d<\infty$ almost surely.

The lifetime of the time-changed process isBy the strong law for Brownian motion, $B_t/t\to 0$ almost surely. If $a<0$, the exponent is eventually at most $at<0$, so the integral is finite. If $a>0$, it is eventually positive and grows linearly; if $a=0$, the recurrence of one-dimensional Brownian motion makes $B$ spend infinite total time in, for example, $[-1,1]$, so the integral is infinite. Consequently