Sigma algebra generated by indicator functions on unit interval I am reading the book Essentials of Probability Theory for Statisticians for self-study and I am stumped on the following exercise (p.47) 2. Let ${(\Omega,\mathcal F, P)}$ be the unit interval equipped with the Borel sets and Lebesgue measure. For $t \in [0,1]$, define $ X_t=  \textbf{1}_t(\omega)$, the indicator function. What is the sigma algebra generated by $ X_t, 0\le t\le1$? Is it the unit interval itself together with the empty set?
 A: Informally, the sigma-algebra generated by a random variable is the smallest sigma-algebra such that this given random variable is measurable with respect to it.
Let me be more precise. Take a random variable $X: \Omega \rightarrow \mathbb{R}$ on a probability space $(\Omega, \mathcal{F}, P)$. The simplest case is when the measurable space we are mapping $\Omega$ into is $(\mathbb{R}, \mathcal{B(\mathbb{R}))}$, the real line equipped with Borel's sigma-algebra.  
Then, we can define the sigma-algebra generated by $X$ writing:
$\mathcal{F}_{X} = \{A \subseteq \Omega : X^{-1}(B) = A \text{, for some borelian set } B \in \mathcal B(\mathbb{R})\}$. 
First, it's easy to see that this collection is a sigma-algebra, indeed, because the inverse image is preserved under countable unions, countable intersections and complementation.
Now, can you see that, by definition, $X$ is $\mathcal{F}_{X}$-measurable? Take some time to convince yourself of this claim. Also, if we take a set out of this collection, it will fail to be a sigma-algebra or $X$ won't be $\mathcal{F}_{X}$-measurable anymore. This is our motivation to call it the sigma-algebra generated by the random variable $X$. This is the smallest collection of sets you need to have in order to say that $X$ is a measurable function.
In your example, $X_{t} = 1, \forall \omega \in \Omega$. So, if you have a given borelian $B$ and $1\in B$, the inverse image is $\Omega$. If not, it's just the empty set and the smallest sigma-algebra needed for "measuring" $X_{t}$ is really just the trivial one, you are right.  
