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Let $(C,\mathcal{F})$ be the set of continous functions $[0,\infty)\to\mathbb{R}$ with the corresponding product $\sigma$-field. Let $Z(f) = \{x\geq0: f(x)=0\}$. Let $\mu$ denote the Lebesgue measure. Denote the collection of closed subsets of $[0,\infty)$ by $D$. Define

$ A = \{f\in C: \mu(Z(f))=0\}.$

I'm interested in understanding the measure-theoretic footing of the statement:

Brownian motion has a zero set of measure 0, almost surely.

If I've interpreted this right, the statement makes sense if $A\in\mathcal{F}$. How do we show that $A\in\mathcal{F}$?

What would be even nicer is if would could find a $\sigma$-field $\mathcal{G}$ on $D$ such that the following holds, with $Z$ and $\mu$ measurable:

$(C,\mathcal{F})\to^Z (D,\mathcal{G})\to^\mu (\mathbb{R}^+,\mathcal{B}),$

where $\mathcal{B}$ is the Borel $\sigma$-field on $\mathbb{R}^+$. This would answer the question of whether $A\in\mathcal{F}$, since the composition of measurable functions is measurable. This is suggested in these notes.

Thanks!

Edit:

By product $\sigma$-field $\mathcal{F}$ on $C$, I mean the smallest $\sigma$-field on $C$ such that, for each $x\geq 0$, the projection map $\pi_x:f\mapsto f(x)$ is measurable. (This definition.)

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The product $\sigma$-field on the set of continuous functions is not particularly useful, since no path is measurable in that $\sigma$-field.

If you look at the problem on a compact interval only, you can take the usual sup-norm and the generated Borel $\sigma$-field. With this norm, the evaluation $e:\mathcal{C}\times[a,b]\to\mathbb{R}$ given by $e(f,x)=f(x)$ is continuous and hence measurable. Therefore, the set $Z=e^{-1}\big(\{0\}\big)$ is measurable. If this set has measure zero, we can conlude by Fubini's theorem that for almost all paths, the set of zeros has Lebesgue measure zero.

Now, if you endow the set $\mathcal{C}$ with the $\sigma$-field generated by the Borel $\sigma$-fields on $C_{[n,n+1]}$ for $n=0,1,\ldots$, you can apply the previous result to conclude again that the set of zeros of almost all paths has Lebesgue measure zero.

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    @NateEldredge I seen now. In that case, the linke given is misleading. Ben uses the trace on $\mathcal{C}$, which does coincide with the Borel $\sigma$-algebra from the topology of uniform convergence.2012-12-03