$\sigma$-algebra generated by random variable I am trying to understand the concept of a $\sigma$-algebra generated by a
random variable. 
Consider the probability space $( [ 0, \infty), \mathcal{F},
P)$. Fix $A \in \mathcal{F}$ such that $P (A) \in (0, 1)$ and let $A^c$ be its complement. Fix $t_0 > 0$ and a
function $$X(t) = 1_{(t_0,\infty)}(t) . 1_A(t)$$
What is the $\sigma$-algebra generated by $X$?
 A: The discussion in the comments to the OP indicates that one is supposed to understand that one considers a single random variable $X$ defined on $Ω=[0,+\infty)$ by $X(t)=\mathbf 1_{t>t_0}\mathbf 1_A(t)$ for every $t$ in $Ω$, or, equivalently, $$X=\mathbf 1_{A\cap(t_0,+\infty)}.$$ User @saz answered that question  perfectly. By contrast, the answer below is not relevant since it addresses the real text of the question, which states that, for every $t$, one considers some random variable $X_t$ defined on $Ω=[0,+∞)$ by $X_t(\omega)=\mathbf 1_{t>t_0}\mathbf 1_A(\omega)$ for every $\omega$ in $Ω$. 
This explains why I deleted my answer, before undeleting it due to a comment by the OP.

Let $X_t=\mathbf 1_{t\gt t_0}\mathbf 1_A$ and $\sigma(X_t)$ the sigma-algebra generated by the random variable $X_t$.


*

*If $t\leqslant t_0$, then $X_t=0$ everywhere hence $\sigma(X_t)=\{\varnothing,\Omega\}$.

*If $t\gt t_0$, then $X_t=\mathbf 1_A$ everywhere and $A\ne\varnothing$, $A\ne\Omega$ hence $\sigma(X_t)=\sigma(A)$, that is, $\sigma(X_t)=\{\varnothing,A,\Omega\setminus A,\Omega\}$.


The process $X=(X_t)_{t\geqslant0}$ is quite degenerate since there exists two random variables $Y$ and $Z$ such that $X_t=Y$ for every $t\leqslant t_0$ and  $X_t=Z$ for every $t\gt t_0$. Thus, the sigma-algebra $\sigma(X)$ generated by the whole process $X$ is $\sigma(Y,Z)$.
In the present case, $Y=0$ and $Z=\mathbf 1_A$ hence $\sigma(X)=\{\varnothing,A,\Omega\setminus A,\Omega\}$.
A: No, that's not correct. Note that for any function $t \mapsto X(t) := X_t$ the $\sigma$-algebra generated by $X$ does not depend on $t$!
Let $B := A \cap (t_0,\infty)$. Then it follows from the definition that $$X_t = 1_B(t).$$ Consequently,
$$\sigma(X) := \sigma(X^{-1}(C); C \, \text{Borel}\} = \{\emptyset,\Omega,B,B^c\} = \{\emptyset,\Omega,A \cap (t_0,\infty), A^c \cup [0,t_0]\}$$
