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I'm proving that if $\mathbf{X}=(X_t)_{t\geq 0}$ is a standard Brownian motion with continuous sample functions, then it holds that the event $(\mathbf{X}\in\mathcal{D})$ is a nullset. Here $\mathcal{D}$ denotes the class of functions that are differentiable in a least one point.

I'm considering the projection $\sigma$-algebra on $\mathbb{R}^{[0, \infty)}$ given by $\mathbb{P}=\sigma\{\hat{\theta}\ |\ \theta\in[0, \infty)\}$. Now could you elaborate the following: $$\big\{x\in\mathbb{R}^{[0,\infty)}\ |\ |x(b)-x(a)|\leq c\big\}=(\hat{a},\hat{b})^{-1}\{(x_1,x_2)\in\mathbb{R}^2\ |\ |x_1-x_2|\leq c\}\in\mathbb{P}$$ Not to get confused, but $x$ is here a function.

How do I get this pre-image to be $\mathbb{P}$-measurable? What is the argument - or the steps, that I'm missing? Thanks.

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    $\begingroup$ What does the hat denote? $\endgroup$ Jan 19, 2020 at 5:15
  • $\begingroup$ It denotes the standard projections, so $\hat{\theta}\,:\,\mathbb{R}^\Theta \rightarrow \mathbb{R}$ $\endgroup$
    – mas2
    Jan 19, 2020 at 16:57

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We may equip $\mathbb{R}^{[0, \infty)}$ with the sigma-algebra generated by the coordinate projections $\hat \theta : \mathbb{R}^{[0, \infty)} \to \mathbb{R}$ mapping $x \mapsto x(\theta)$. This is the smallest sigma-algebra which makes all projection mappings $\mathbb P/ \mathcal B(\mathbb R)$-measurable, i.e. the smallest $\sigma$-algebra which contains $${\hat \theta}^{-1}(A) = \{ x \in \mathbb{R}^{[0, \infty)} \mid \hat\theta(x) \in A\} = \{ x \in \mathbb{R}^{[0, \infty)} \mid x(\theta) \in A\}$$ for all $\theta \in [0, \infty)$ and all $A \in \mathcal{B}(\mathbb{R})$.

Now let $A = \{(x_1, x_2) \in \mathbb{R}^2 \mid |x_1 - x_2| \leq c \}$ and note that $A \in \mathcal B(\mathbb R^2)$. Since the bundling $(\hat a, \hat b)$ of two $\mathbb P/\mathcal B(\mathbb R)$-measurable functions is $\mathbb P/\mathcal B(\mathbb R^2)$-measurable, we find that the preimage of $A \in \mathcal B(\mathbb R^2)$ under $(\hat a, \hat b)$ is a $\mathbb{P}$-measurable set, i.e. $$ \{x \in \mathbb{R}^{[0, \infty)} \mid |x(a) - x(b)| \leq c \} = \{x \in \mathbb{R}^{[0, \infty)} \mid (\hat a, \hat b)\circ x \in A \} = (\hat a, \hat b)^{-1} (A) \in \mathbb{P} .$$ So in short, your set is $\mathbb P$-measurable because it is the preimage of a Borel-set under a projection, and $\mathbb P$ is defined as precisely the sigma-algebra making the projection mappings measurable.

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  • $\begingroup$ Thanks. This was exactly what I was looking for! Can you then explain why the set $$\mathcal{D}=\bigcup_{t\geq0} \{x\in\mathbb{R}^{[0,\infty)}\ |\ x\ \text{is differentiable in}\ t\}$$ might not be $\mathbb{P}$-measurable? I know it's an uncountable union since $t\in[0,\infty)$, but the event, how might this not be $\mathbb{P}$-measurable? $\endgroup$
    – mas2
    Jan 20, 2020 at 0:46
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    $\begingroup$ First, even if each set $\mathcal D^t = \{x \in \mathbb R^{[0, \infty)} \mid x \text{ is differentiable in } t \}$ is measurable, this does not imply that the uncountable union $\mathcal D = \bigcup_{t \geq 0} \mathcal D^t$ is also measurable. But even the individual sets $\mathcal D^t$ seem unlikely to be measurable -- differentiability is a rather delicate property, and it is likely not captured by the preimage of some Borel set. I don't have a proof however, but it is intuitively similar to the non-measurability of $C([0, \infty))$. $\endgroup$ Jan 20, 2020 at 9:30

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