Does zero variance of a Riemann integral imply constant integrand?

Question

Let $f(s)$ be a function adapted adapted to the filtration of Brownian motion. Consider the integral $$X=\int_0^T f^2(s)ds$$ If $\operatorname{Var}(X)=0$ then is $f$ deterministic? I know $X$ is, but what if $f$ has some cancellation built into it?

Answer 1

$f$ need not be deterministic. For example let $Y$ be an $\mathcal{F}_0$-measurable random variable with $$\mathbb{P}(Y = 1) = \mathbb{P}(Y = -1) = \frac12$$ where $(\mathcal{F}_t)$ is the Brownian filtration.

Then define $f$ by $$f(\omega, t) = \begin{cases} 1 - Y(\omega) \, \, \text{ if } t < \frac12 \\ 1 + Y(\omega) \, \, \text{ otherwise } \end{cases}$$ Then if $T = 1$, $X(\omega) = 1$ for any $\omega$ but for each fixed $t$, $$\mathbb{P}(f(\cdot, t) = 2) = \mathbb{P}(f(\cdot, t) = 0) = \frac12.$$