The following are some problems I encountered when self-learning GTM 261 "Probability and Stochastics".
Definition (determinability)
If $X$ and $Y$ are random variables taking values in $(E,\mathcal{E})$ and $(D,\mathcal{D})$, then we say that $X$ determines $Y$ if $Y=f\circ X$ for some $f:E\rightarrow D$ measurable with respect to $\mathcal{E}$ and $\mathcal{D}$.
Problem 1
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 & \text{if } t < T(\omega) \\ 1 & \text{if } t \geq T(\omega) \end{cases} $ Show that $X$ and $T$ determine each other. If $T$ represents the time of failure for a device, then $X$ is the process that indicates whether the device has failed or not. That $X$ and $T$ determine each other is intuitively obvious, but the measurability issues cannot be ignored altogether.
In particular, I do not know how to show the measurability part.
Problem 2
A slight change in the preceding exercise shows that one might guard against raw intuition. Let $T$ have a distribution that is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}_+$; in fact, all we need is that $\mathbb{P}\{T = t\} = 0$ for every $t\in\mathbb{R}_+$. Define $ X_t(\omega) = \begin{cases} 1 & \text{if } t = T(\omega)\\ 0 & \text{otherwise} \end{cases} $ Show that, for each $t\in\mathbb{R}_+$, the random variable $X_t$ is determined by $T$. But, contrary to raw intuition, $T$ is not determined by $X=(X_t)_{t\in\mathbb{R}_+}$. Show this by the following steps below:
a. For each $t$, we have $X_t = 0$ almost surely. Therefore, for every sequence $(t_n)$ in $\mathbb{R}_+$, $X_{t_1} = X_{t_2} = \ldots = 0$ almost surely.
b. If $V\in \sigma(X)$, then $V = c$ almost surely for some constant $c$. It follows that $T$ is not in $\sigma(X)$.