I have this exercise (Cinlar, Ch.2 ex. 4.14 ):
Let $T$ be a positive random variable and define a stochastic process $X = (X_t)_{t \in \mathbb{R}_+}$ by setting, for each $\omega$ $$ X_t(\omega) = \begin{cases} 0 & t< T(\omega) \\ 1 & t\ge T(\omega) \end{cases} $$ Show that $X$ and $T$ determine each other.
So I have to find two measurable functions $f,g$ such that \begin{align*} X &= f \circ T \\ T &= g \circ X \end{align*}
the first one is easy. I have some problems with the second: \begin{align*} \varphi: 2^{\mathbb{R}_+} &\longrightarrow \mathbb{R}_+ \\ x &\longmapsto \inf\{t \in \mathbb{R}_+ : x_t = 1\} \end{align*}
I have that $T = \varphi \circ X$, so I'd like to say that $\varphi$ is measurable. I have that $$\varphi^{-1}(\alpha,+\infty) = \{x \in 2^{\mathbb{R}_+} : x_t = 0 \; \forall t\in[0, \alpha] \}$$ If what I've said till now is right, how can I show that this set is measurable in the product space? Thanks
EDIT: as pointed out in the comments this set is not measurable in $2^{\mathbb{R}_+}$. But the deterministic function $\varphi$ needs to have as domain the measurable space associated to the stochastic process $X$, which is $2^{\mathbb{R}_+}$ with its product $\sigma$-algebra... How can I solve this?
It can be shown that $T$ is determined by $(X_t)_{t \in \mathbb{Q}_+}$ (and a fortiori by $(X_t)_{t \in \mathbb{R}_+}$). I take the same function $\varphi$, but now I define it over $2^{\mathbb{Q}_+}$ \begin{align*} \varphi: 2^{\mathbb{Q}_+} &\longrightarrow \mathbb{R}_+ \\ x &\longmapsto \inf\{t \in \mathbb{Q}_+ : x_t = 1\} \end{align*}
Now, it is evident that $T = \varphi \circ (X_t)_{t \in \mathbb{Q}_+}$, since $\mathbb{Q}_+$ is dense in $\mathbb{R}_+$, it remains to show that $\varphi$ is $\mathscr{P}(\{0,1\})^{\mathbb{Q}_+}$-measurable.
But this is true because, in fact, since $\mathbb{Q}_+$ is countable, we have that $\mathscr{P}(\{0,1\})^{\mathbb{Q}_+} = \mathscr{P}(\{0,1\}^{\mathbb{Q}_+})$, and therefore every numerical function over $2^{\mathbb{Q}_+}$ is measurable.