Given $\mu_t$ continuous stochastic process that satisfies $\int_0^t \mu_s^2\;ds<\infty$. Define $X_t\equiv \int_0^t \mu_s\;ds$. Let $|\cdot|$ denote floor function. Then where does $\sum_{n=1}^{|t/\Delta|}(X_{n\Delta}-X_{(n-1)\Delta})^2$ converge to in Mean Square as $\Delta\rightarrow 0$?
This is not HW. My attempt has been to write $\sum_{n=1}^{|t/\Delta|}(X_{n\Delta}-X_{(n-1)\Delta})^2$ as $\sum_{n=1}^{|t/\Delta|} (\int_{(n-1)\Delta}^{n\Delta} \mu_s\;ds)^2$ and then I'm not sure how to proceed, despite knowing the definition of MS convergence.
