I am reading a textbook and I am stuck proving a seemingly trivial point (but I believe I am not the only one that was pondering this question and it shows similarity to this question). Let me sketch the setting: let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space endowed with a filtration $\mathbb{F}:=\{F_{t}:t\in[0,T]\}$. We assume that the filtration satisfies the usual conditions (and that $\mathcal{F}$ is complete). Let $E$ be a separable Banach space and let $[0,T]\subset\mathbb{R}$ be a finite interval equipped with the Borel $\sigma$-algebra.
We have an $E$-valued progressively measurable process $Y=\{Y_{t}:t\in[0,T]\}$ that is square-integrable on $[0,T]\times\Omega$. The textbook I'm reading claims that the process $\int_{0}^{\cdot}Y(s)\text{ d}s:=\left\{\int_{0}^{t}Y(s)\text{ d}s:t\in[0,T]\right\}$ is continuous and adapted.
The adaptedness is not the issue: we can invoke Fubini's theorem (for Bochner integrals) and use the progressive measurability (and integrability) of $Y$ to obtain that the map $\omega\mapsto \int_{0}^{t}Y(s,\omega)\text{ d}s$ is indeed measurable with respect to $\mathcal{F}_{t}$ for every $t\in[0,T]$. However, the part that nags me is the continuity.
Since $Y$ is square-integrable on $[0,T]\times\Omega$, we know that $$ \mathbb{E}\left[\int_{0}^{T}\left\|Y(s)\right\|_{E}^{2}\text{ d}s\right]<\infty, $$ meaning that on a (measurable) set $\Omega_{Y}$ of full probability, we have $\int_{0}^{T}\left\|Y(s)\right\|_{E}^{2}\text {d}s<\infty$, implying that $\int_{0}^{T}\left\|Y(s)\right\|_{E}\text {d}s<\infty$, as $[0,T]$ is a finite measure space. Then, using the Bochner inequality. we see that the map $t\mapsto \int_{0}^{t}Y(s,\omega)\text{ d}s$ is continuous in $E$ for all $\omega\in\Omega_{Y}$.
Right now, I am stuck: I can not guarantee that all sample paths are continuous. I see that the process $\int_{0}^{\cdot}Y(s) \text{ d}s$ is indistinguishable from the process $\textbf{1}_{\Omega_{Y}}\int_{0}^{\cdot}Y(s) \text{ d}s := \left\{\textbf{1}_{\Omega_{Y}}\int_{0}^{t}Y(s) \text{ d}s:t\in[0,T]\right\}$ and the latter process is adapted and is continous, but I do not see how to show that $\int_{0}^{\cdot}Y(s) \text{ d}s$ is continuous.
Thanks in advance!