# Function evaluated in Brownian motion vanishes implies that the function itself vanishes?

tl;dr, here's my question:

Question. Let $f(t,x)$ be a measurable function such that $f(t,B_t)=0$ almost everywhere on $[0,T]\times\Omega$ for a Brownian motion $B_t$. Does this imply that $f(t,x)=0$ almost everywhere on $[0,T]\times\mathbb R$?

Given that Brownian motion is recurrent and normally distributed (and hence, the probability that $|B_t|>a$ for any $a$ is nonzero, albeit very small for large $a$), it seems intuitive enough that this should be true, but I have been unable to prove it conclusively.

For context: I was reading the answer to this question (which can actually be found on this MO question), and there was a detail that left me unsatisfied.

If we remove any specificities related to the linked problem and look at the core of the argument, unless I'm mistaken, it goes as follows (tl;dr skip to question below):

1. Let $(B_t)_{t\geq0}$ be a Brownian motion (for simplicity), and let $f(t,x):\mathbb R^+\times\mathbb R\to\mathbb R$ be a measurable function.
2. Suppose that $\int_0^tf(s,B_s)~\text d s=0$ for all $t\leq T$. Then, for almost every $\omega$, $f\big(t,B_t(\omega)\big)=0$ for almost every $t$ (by the Lebesgue differentiation theorem).
3. By Fubini's theorem, we can argue that $f\big(t,B_t(\omega)\big)=0$ almost everywhere on $[0,T]\times\Omega$.
4. Therefore $f(t,x)=0$ for almost all $(t,x)$.

The only part of the argument that I'm not confortable with (which is the one that is left unexplained in the linked MO answer) is part 4, which is essentially the question above.

Let $D:=\{(t,\omega):f(t,B_t(\omega)\not=0\}$ and $C=\{(t,x):f(x,t)\not=0\}$. Observe that $D=\varphi^{-1}(C)$, where $\varphi(t,\omega) =(t,B_t(\omega))$ (mapping $[0,T]\times\Omega$ into $[0,T]\times\Bbb R$). Consequently, writing $\lambda$ for Lebesgue measure on $[0,T]$, $$0=\lambda\otimes\Bbb P(D) =\int\int 1_{C}(t,x){1\over\sqrt{2\pi t}}e^{-x^2/2t}\,dt\,dx.$$ Because the integrand $(t,x)\mapsto{1\over\sqrt{2\pi t}}e^{-x^2/2t}$ is strictly positive, it must be that $\int\int 1_{C}(t,x)\,dt\,dx=0$. That is, $\lambda\otimes\Bbb P(C)=0$.

For any measurable set $A \subseteq [0,T] \times \Omega$ it holds that

\begin{align*} \{(t,\omega); (t,B_t(\omega)) \in A\} &= \int_0^T \! \int 1_A(t,B_t) \, d\mathbb{P} \, dt \\ &= \int_0^T \! \int 1_A(t,x) \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t} \, dx \, dt. \end{align*}

Since $e^{-x^2/2t}$ is strictly positive, this implies

$$(\lambda|_{[0,T]} \times \mathbb{P})(A) = 0 \iff (\lambda|_{[0,T]} \times \mathbb{P})(\{(t,B_t) \in A\})=0.$$

Consequently, if we set $A:= \{(t,x); f(t,x) \neq 0\}$, then we find that

$$(\lambda|_{[0,T]} \times \mathbb{P})(\{(t,\omega); f(t,B_t(\omega)) \neq 0\}) = 0$$

if, and only if,

$$(\lambda|_{[0,T]} \times \mathbb{P})(\{(t,x); f(t,x) \neq 0\})=0.$$