Expectation Conditioned on $\sigma$-subalgebra

Question

In preparing for my upcoming qualifying exam, I have encountered the following problem:

Consider the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ where $\Omega = [0,1]$, $\mathcal{F}$ is the Borel algebra on $\Omega$, and $\mathbb{P}$ is the Lebesgue measure. If $X(\omega) = \max{(\sin{(2\pi \omega)}, 0)}$ and $Y(\omega) = X^2(\omega)$, find the $\sigma$-subalgebra generated by random variable $X$, $\sigma(X)$, and $\text{E}[Y \vert \sigma(X)]$.

The second part of this problem seems quite a bit easier than the first, i.e. finding $\text{E}[Y \vert \sigma(X)]$. Since $Y$ is a function of $X$, random variable $Y$ must be measurable with respect to $\sigma(X)$ and so

$$ \text{E}[Y \vert \sigma(X)] = \text{E}[X^2 \vert \sigma(X)] = \text{E}[X^2 \vert X] = X^2 $$

I have difficulty finding an explicit expression for $\sigma(X)$, however. I understand that for $\omega \in (0,0.5)$ we have $X(\omega) = \sin{(2\pi\omega)}$ and for $\omega \in \{0\}\cup[0.5,1]$ we have $X(\omega)$. This leads me to write

$$ \sigma(X) = \{ \varnothing, \{X \neq 0\}, \{X = 0\} ,\Omega\} = \{ \varnothing, (0,0.5), \{0\}\cup[0.5,1] ,\Omega\} $$

However, I am not very confident in this answer. As a general definition, $\sigma(X)$ is the smallest $\sigma$-algebra on the defined probability space that contains all sets of the form $\{X (\omega) \leq a\}$ for all $a \in \mathbb{R}$. For $a < 0$, this is represented by the $\varnothing$ element, and for $a = 0$ we have $\{0\}\cup[0.5, 1]$, and for $a \geq 1$ we have $\Omega$. For $a \in (0,1)$, should the set be of the form

$$ \{\omega \in \Omega \colon X(\omega) \leq a\} = [0, \tfrac{\arcsin{(a)}}{2\pi}] \cup [0.5 - \tfrac{\arcsin{(a)}}{2\pi}, 1] $$

and then this would replace the $(0,0.5)$ element in the above $\sigma(X)$ expression? Perhaps something of the form

$$ \bigcup_{a \in (0,1)} \left( [0, \tfrac{\arcsin{(a)}}{2\pi}] \cup [0.5 - \tfrac{\arcsin{(a)}}{2\pi}, 1] \right) $$

Am I allowed to have a function of another variable $a$ in my representation of $\sigma(X)$? Any guidance is greatly appreciated!

Answer 1

Yes, it is certainly fine to have another variable to index the union in your expression for $\sigma(X)$. The tricky part about trying for such an explicit expression for $\sigma(X)$ is that it's difficult to make it closed under unions. Instead, it's easier to just work from the Borel $\sigma$-algebra. Since $X$ is invertible on $(0,\frac 14)$, and $X(\frac 12 - \omega) = X(\omega)$ for $\omega \in (0,\frac 14)$, we have that if $A \in \mathcal{F}|_{(0,\frac 14)}$ then $A \cup (\frac 12 - A) \in \sigma(X)$. Basically, if we know $X$, we can determine whether or not $\omega$ is in either $A$ or $\frac 12 - A$. Now, if $X = 0$, then $\omega \in \{0\} \cup [\frac 12,1]$. Hence,

\begin{align*} \sigma(X) &= \left\{\{0\} \cup [\frac 12,1],A \cup (\frac 12 - A) : A \in \mathcal{F}|_{(0,\frac 14)}\right\}. \end{align*}