# What does it mean by saying 'a random variable $\mathit X$ is $\mathcal G$-measurable'?

This is the definition of "measurability":

Let $$\mathit X$$ be a random variable defined on $$(\Omega,\mathcal F,\mathbb P)$$. Let $$\mathcal G$$ be a $$\sigma$$-algebra of subsets of $$\Omega$$. If $$\sigma(\mathit X) \subseteq \mathcal G$$, we say that $$\mathit X$$ is $$\mathcal G$$-measurable.

I can understand this definition, but what does it mean that "a random variable $$\mathit X$$ is $$\mathcal G$$-measurable if and only if the information in $$\mathcal G$$ is sufficient to determine the value of $$\mathit X$$"?

My questions are:

1. What is the "information" in $$\mathcal G$$? As far as I know, $$\mathcal G$$ is just a set of subsets of $$\Omega$$. Thus $$\mathcal G$$ contains lots of "events". Then, when we say "information", what exactly do we mean？ I can't imagine other information except for the elements of $$\mathcal G$$.
2. How can the "information" in $$\mathcal G$$ determine the value of $$\mathit X$$? Given $$\mathcal G$$, what we know is just "the elements of $$\mathcal G$$". How could this "infromation" help us determine the value of $$\mathit X$$?

Example:

This is an example of rolling a die with $$\Omega = \{1, 2, 3, 4, 5, 6\}$$:

$$\mathit X_1(\omega)=\omega$$

$$\mathit X_2(\omega)=\begin{cases} 1, & \omega\in \{1,3,5\} \\-1, & \omega\in \{2,4,6\}\end{cases}$$

$$\mathit X_1$$ and $$\mathit X_2$$ are both random variables. $$\mathit X_1$$ gives the exact outcome of the roll, and $$\mathit X_2$$ is a binary variable whose value depends on whether the roll is odd or even.

Let $$\mathcal G = \{\emptyset, \Omega, \{1,3,5\}, \{2,4,6\}\}$$, then $$\mathit X_2$$ is measurable w.r.t $$\mathcal G$$ but $$\mathit X_1$$ is not measurable w.r.t $$\mathcal G$$. I know this because I can check that $$\sigma(X_2)=\mathcal G$$ according to the defination of $$\sigma(X_2)$$.

Here are the questions:

1. What's the information in $$\{\emptyset, \Omega, \{1,3,5\}, \{2,4,6\}\}$$? What else can we know except for those four elements in $$\{\emptyset, \Omega, \{1,3,5\}, \{2,4,6\}\}$$?

2. If this is the only 'information' that we can obtain from $$\mathcal G$$ (i.e., there are four elements in $$\mathcal G$$, and those elements are $$\emptyset$$, $$\Omega$$, $$\{1,3,5\}$$ and $$\{2,4,6\}$$), how can we determine what the value of $$\mathit X_2$$ will be?

• What is your definition of $\sigma$? Commented May 23, 2020 at 6:13
• This intuition about 'the information in G' is clarified by thinking about $G$ as being $G = \sigma( Y_1, \ldots, Y_m)$, where the $Y_i$ are measurements that someone is making about some experiment for which $X$ is another possible measurement. If $X \in \sigma(Y_1, \ldots, Y_m)$, you can express the value of $X$ in terms of the values of $Y_i$ (literally there is some measurable function $f$ so that $X = f(Y_1, \ldots, Y_n$) ) -- hence you can express $X$ in terms of the information in the $Y_i$ which is to say, the information in $G$. Commented May 23, 2020 at 6:17
• @DEATH_CUBE_K Hi, $\sigma$-algebra is the subset of $\Omega$ that includes $\Omega$ itself, is closed under complement, and is closed under countable unions. I can understand this concept. Commented May 23, 2020 at 6:18
• Saying $\sigma(X) \subseteq G$ is the same as saying that $X$ is $G$ measurable, because it is asserting that all events $\{X \leq a \}$ are in $G$. Commented May 23, 2020 at 6:21
• "I still can't understand what information is in G except for knowing the elements of it. " -- if you want, you can think about the 'information' in $G$ as observations about which events in $G$ occured in our experiment: We get some example $\omega$ from our probability distribution, and then the information in G is the information of, for each $A \in G$, whether $\omega \in A$ (that is, whether $A$ occurred in the universe/outcome $\omega$). Commented May 23, 2020 at 6:49