Consider the SDE
\[
dX(t)=b(X(t))dt +\sigma(X(t))dB(t)
\]
with \(b(x)\to b_0 >0\) as \(x\to\infty\) and with \(\sigma\) bounded and positive. Suppose that \(b\) and \(\sigma\) are such that
\[\lim_{t\to\infty}X(t)=\infty\], with probability one for any starting point. Show that
\[
P_x\Big\{\lim_{t\to\infty}\frac{X(t)}{b_0 t}=1\Big\}=1 \ .
\]
From
\[
X(t)=x+\int_0^{t}b(X(s))ds +\int_0^{t}\sigma(X(s))dB(s)
\]
and the hypotheses, note that the result follows from showing that
\begin{align*}
\mathbf P_x\Big\{\lim_{t\to\infty}\frac{1}{t}\int_0^{t}\sigma(X(s))dB(s)=0\Big\}=1 \ .
\end{align*}

There are a number of ways of thinking about this. In the end they all come down to essentially the same calculations. One way is to show that for some fixed \(\delta \in(0,1)\) the following statement holds with probability one:

There exist a constants \(C(\omega)\) so that
\begin{align*}
\int_0^{t}\sigma(X(s))dB(s) \leq Ct^\delta
\end{align*}
for all \(t >0\).

To show this partition \([0,\infty]\) into blocks and use the Doob-Kolmogorov inequality to estimate the probability that the max of \( \int_0^{t}\sigma(X(s))ds\) on each block excess \(t^\delta\) on that block. Then use the Borel-Cantelli to show that this happens only a finite number of times.

A different way to organize the same calculation is to estimate
\[
\mathbf P_x\Big\{\sup_{t>a}\frac{1}{t}|\int_0^t \sigma(X(s))dB(s)|>\epsilon\Big\}
\]
by breaking the interval \(t>a\) into the union of intervals of the form \(a2^k <t\leq a2^{k+1}\) for \(k=0,1,\dots\) and using Doob-Kolmogorov Martingale inequality. Then let \(a\to\infty\).