Is $\lvert \int_0^t e^{isB_s} ds \rvert^2 \leq \left( \int_0^t \lvert e^{isB_s}\rvert ds\right)^2 = \left( \int_0^t 1 ds \right)^2 = t^2$ correct?

Question

Assuming $B$ is a standard Brownian motion and taking the Riemann integral as below, where $i = \sqrt{-1}$, is there anything wrong with this inequality?

$\lvert \int_0^t e^{isB_s} ds \rvert^2 \leq \left( \int_0^t \lvert e^{isB_s}\rvert ds\right)^2 = \left( \int_0^t 1 ds \right)^2 = t^2$.

I'm just confused, since $B_s$ is a random variable, but does that matter? Assuming the first ''triangle inequality'' holds: $e^{isB_s}$ is a random variable, but it's still a complex exponential, whose modulus is equal to $1$, or? I mean, for any $\omega$, $\lvert e^{isB_s(\omega)} \rvert^2 = 1$, right?

Answer 1

Yes, this is perfectly fine. All you're using is that, for each $\omega$ in the sample space, $B_s(\omega)$ is a real-valued measurable function of $s$.