This question is from Problem 8.2 and its solution (Schilling's "Brownian Motion"). I cannot understand following proof.
Problem 8.2
Let $B_t$ be a d-dimensional Brownian motion, $f:\mathbb{R}^d\to\mathbb{R}$ be a continuous function such that $\int_0^tf(B_s)ds=0$ for all $t>0$. Show that $f(B_s)=0$ for all $s>0$, and conclude that $f\equiv0$.
proof
By assumption, $f(B_t)=\frac{d}{dt}\int^t_0f(B_s)ds=0$. We can assume that $f$ is positive and bounded, otherwise we could consider $f^{\pm}(B_t)\wedge c$ for some constand $c>0$. Now $E[f(B_t)]=0$ and we conclude from this that $f=0$.
In the cases of positive and bounded function, I can understand that $E[f(B_t)]=0$ implies $f\equiv0$. But, I cannot understand "otherwise we could consider $f^{\pm}(B_t)\wedge c$ for some constand $c>0$" part.
For example, suppose $f$ is bounded but is not positive. In this case, I get $E[f^+]-E[f^-]=0$ by $f(B_t)=0$. How can I get $f\equiv0$ from this equation? And, the case of $f$ which is neither bounded nor positive, why $f\equiv0$?
Thank you for your cooperation.
The highlighted proof is not valid. The argument is much simpler. The first step in the proof shows that $f(B_1)=0$ and this is enough to conclude that $f \equiv 0$. This is because the distribution of $B_1$ has full support: If $f \neq 0$ on set $E$ of positive measure then $P(B_1 \in E) >0$ and this leads to a contradiction to $f(B_1)=0$. So $f=0$ a.e. which implies $f \equiv 0$ by continuity.