In the book "Introduction to Stochastic Integration" by Kuo, at page $46$, he has written:
$\int_{a}^{b} E(|f(t)-g_{n}(t)|^{2})dt \\ \leq \int_{a}^{b} \int_{0}^{\infty} e^{-\tau }E(|f(t)-f(t-\frac{\tau}{n})|^{2})d\tau dt\\ = \int_{0}^{\infty} e^{-\tau }\int_{a}^{b} E(|f(t)-f(t-\frac{\tau}{n})|^{2}) dt d\tau\\ = \int_{0}^{\infty} e^{-\tau }E(\int_{a}^{b} |f(t)-f(t-\frac{\tau}{n})|^{2} dt )d\tau $
where: $g_{n}(t,\omega) := \int_{0}^{n(t-a)}e^{-\tau}f(t- \frac{\tau}{n}, \omega)d\tau$ ; $f$ is BOUNDED ; AND assuming $f(t) = 0 ; \forall t \lt a$ it has been justified that: $f(t) - g_{n}(t) = \int_{0}^{\infty}e^{-\tau}[f(t) - f(t-\frac{\tau}{n})]d\tau$.
After this author has written that:(in $4.3.11$)
Since $f$ is assumed to be bounded we have: $\int_{a}^{b} |f(t,.)-f(t-\frac{\tau}{n},.)|^{2} dt \to 0$ almost surely, as $n \to \infty$.
Now, my question is: how this almost sure convergence is justified??? [SINCE: $f$ is assumed to be BOUNDED only, NOT CONTINUOUS!! Can someone explain it to me please?? Thank You!!