# Projection measurability of a class of functions

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.

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

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.
• 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?
• 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 $C([0, \infty))$. Jan 20, 2020 at 9:30