Two questions on determinability (probability theory) 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 $\sigma(X)$.

 A: Problem 1: Fix $t\in\mathbb{R}_+$. Then $X_t=1_{\{T\le t\}}$. Since $\{T\le t\}\in\sigma T$, it follows that $X_t$ is $\sigma T$-measurable. Therefore, $\sigma X_t\subset\sigma T$, and this implies
  $$
  \sigma X = \bigvee_{t\in\mathbb{R}_+}\sigma X_t \subset\sigma T.
  $$
By Theorem 4.4, $X$ is a measurable function of $T$, and so $T$ determines $X$.
For the converse, fix $t\in\mathbb{R}_+$. Then
  $$
  \{T\le t\} = \{X_t = 1\} \in \sigma X_t \subset \sigma X.
  $$
Since $t$ was arbitrary, this gives $\sigma T\subset\sigma X$, and so $X$ determines $T$.
Problem 2(a): For each $t$, we have $P(X_t \ne 0) = P(T = t) = 0$. Hence, $X_t=0$ a.s. Moreover, given a sequence $(t_n)$,
  $$
  P(\exists n\text{ such that }X_{t_n}\ne 0)
    = P\bigg(\bigcup_{n=1}^\infty\{X_{t_n}\ne 0\}\bigg)
    \le \sum_{n=1}^\infty P(X_{t_n}\ne 0) = 0.
  $$
Thus, $P(X_{t_n}=0,\forall n)=1$, that is, $X_{t_1}=X_{t_2}=\cdots=0$ a.s.
Problem 2(b): Suppose $T$ is $\sigma X$ measurable. Then by Proposition 4.6, we have
  $$
  T = f(X_{t_1},X_{t_2},\ldots),
  $$
for some sequence $(t_n)$ and some Borel-measurable function $f:\mathbb{R}^\infty \to \mathbb{R}_+$. Define $t=f(0,0,\ldots)\in\mathbb{R}_+$. Then, by part (a), we have
  $$
  T = f(X_{t_1},X_{t_2},\ldots) = f(0,0,\ldots) = t\text{ a.s.}
  $$
But this contradicts the hypothesis that $P(T=t)=0$ for all $t$.
