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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.)

Glorfindel
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Ben Derrett
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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.

Community
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Michael Greinecker
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  • Sorry Michael. I was unclear using the term product $\sigma$-field. I've edited my post to clarify exactly which $\sigma$-field I mean. Paths should now be measurable. – Ben Derrett Dec 03 '12 at 11:28
  • @BenDerrett I is quite clear what you mean by product $\sigma$-field, but there is no sample path $f$ such that ${f}$ is measurable in the product $\sigma$-field. – Michael Greinecker Dec 03 '12 at 12:41
  • Thanks. What about this: Let $f$ be the element of $C$ which is identically $0$. Then we have ${f}=\cap_{q\in\mathbb{Q}^+}\pi_q^{-1}(0)\in\mathcal{F}$. – Ben Derrett Dec 03 '12 at 13:15
  • @BenDerrett No. The indicator function of the positive irrationals is also in the intersection. The product $\sigma$-field doesn't contain just continuous functions and the set of continuous functions is not a measurable subset. – Michael Greinecker Dec 03 '12 at 14:02
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    @MichaelGreinecker: The indicator of the irrationals is not in $C$. We are only defining a $\sigma$-field on $C$ so certainly $C$ is measurable. – Nate Eldredge Dec 03 '12 at 14:39
  • @BenDerrett: In fact, both you and Michael are talking about the same $\sigma$-field. But either way, the essential point is that the measurability of the set $A$ comes from Fubini's theorem. – Nate Eldredge Dec 03 '12 at 14:42
  • @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. – Michael Greinecker Dec 03 '12 at 15:04