Let $(B_t)_{t\ge 0}$ be a BM$^d$, $f: \mathbb{R}^d \to \mathbb{R}$ be a continuous function such that $\int_0^t f(B_s)ds=0$ for all $t>0.$ Show that $f(B_s)=0$ for all $s>0$, and conclude that $f\equiv 0$.
This is a problem from Rene Schilling's Brownian motion and we have the following solution.
By differentiation we get $\frac{d}{dt}\int_0^t f(B_s)ds=f(B_t)$ so that $f(B_t)=0$. We can assume that $f$ is positive and bounded, otherwise we could consider $f^{\pm}(B_t) \wedge c$ for some constant $c>0$. Now $E f(B_t)=0$ and we conclude from this that $f=0$.
My question is: How do we conclude from $Ef(B_t)=0$ that $f=0$? The expectation of a positive measurable function being $0$ implies that the function is a.s. $0$. But here the function is $f(B_t)$ not just $f$.
If $f$ is not identically $0$, there is some $x$ such that $f(x) \ne 0$. WLOG say $f(x) > 0$. By continuity there is $\epsilon > 0$ such that $f(y) > 0$ for $\|y - x\| < \epsilon$. Now for any $t > 0$, $\mathbb P(\|B_t - x\| < \epsilon) > 0$, so $\mathbb P(f(B_t) > 0) > 0$.