Return to Revisions

Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. First by (uniform) continuity in the $L^2$ norm, for each fixed $T > 0$ and integer $n > 0$, there exists a $\delta_n > 0$ such that $$ \mathbb{E}\left[(X(t)-X(s))^2\right] \le 2^{-n} $$ for all $0\le s\le t\le T$ with $\lvert s-t\rvert\le\delta_n$. Choose partitions $0=t^n_0\le t^n_1\le\cdots\le t^n_{k_n}=T$ with mesh $\max_i(t^n_i-t^n_{i-1})\le\delta_n$. As in the question, define the simple functions $$ \xi_n(t) = \sum_{i=0}^{k_n-1}X(t_i^n)1_{\lbrace t_i^n \le t < t_{i+1}^n\rbrace} $$ so that, again as mentioned in the question, the Bochner integral is given by $$ \int_0^T\xi_n(t)\\,dt \rightarrow\text{(B-)}\int_0^T X(t)\\,dt. $$ Here the limit is taken in the $L^2$ norm and, hence, also holds for convergence in probability. However, each $t\in[0,T)$, lies in an interval $[t^n_i,t^n_{i+1})$ with $t-t^n_i\le\delta_n$, so $$ \mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]=\mathbb{E}\left[\lvert X(t)-X(t^n_i)\rvert\right]\le2^{-n/2}. $$ Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. $$ \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\lvert X(t)-\xi_n(t)\rvert\\,dt\right] &=\int_0^T\sum_{n=1}^\infty\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\\,dt\cr &\le\int_0^T\sum_{n=1}^\infty2^{-n/2}\\,dt\cr &=T/(2^{1/2}-1) < \infty. \end{align} $$ In particular, $$ \int_0^T\sum_{n=1}^\infty\lvert X(t)-\xi_n(t)\rvert\\,dt < \infty $$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$ \int_0^T\xi_n(t)\\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\\,dt. $$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.