# Independence vs. Measurability

As someone who is relatively new to measure-theoretic probability, I'm really struggling to get to grips with the concepts of measurability and independence.

If we have a random variable $$X$$ defined on a probability space $$(\Omega,\mathscr{F},\mathbb{P})$$ and some sigma algebra $$\mathscr{G}\subset\mathscr{F}$$, and $$X$$ is independent of $$\mathscr{G}$$, then can we deduce that $$X$$ is not $$\mathscr{G}$$-measurable?

When I look at the definitions of measurability and independence they seem pretty distinct from one another, but when heuristically considering sigma algebras as "information", it seems like if our random variable is independent of our information (i.e. $$X$$ independent of $$\mathscr{G}$$), then we don't have enough information to determine the value of our random variable (i.e. $$X$$ is not $$\mathscr{G}$$-measurable).

If someone could please sort out my confusion it would be much appreciated!

## 2 Answers

At least assuming that $$X$$ is non-trivial, you are correct that $$X$$ is independent of $$\mathscr G$$ implies $$X$$ is not $$\mathscr G$$-measurable. For a proof, suppose to the contrary that $$X$$ is independent of $$\mathscr G$$ and $$X$$ is $$\mathscr G$$-measurable. Then $$\mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B)$$ for all $$A \in \sigma(X)$$ and $$B \in \mathscr G$$. But since $$X$$ is $$\mathscr G$$-measurable, $$\sigma(X) \subseteq \mathscr G$$ so we can choose $$B = A$$ and hence have $$\mathbb P(A) = \mathbb P(A \cap B) = \mathbb P(A) \mathbb P(B) = \mathbb P(A)^2$$ for all $$A \in \sigma(X)$$. But $$\mathbb P(A) = \mathbb P(A)^2$$ implies $$\mathbb P(A) = 0$$ or $$\mathbb P(A)=1$$, so $$X$$ would have to be trivial.

The other implication is not true, i.e. $$X$$ is not $$\mathscr G$$-measurable does not imply $$X$$ is independent of $$\mathscr G$$. You could consider a simple example of tossing a fair coin twice, $$X$$ is the total number of heads, and $$\mathscr G$$ is the $$\sigma$$-algebra generated by the first coin toss.

• This is really clear - thank you so much! :) Commented Feb 3, 2021 at 16:50

Assuming that X is non-trivial, we can find that $$E[X|\mathcal{G}]=EX$$, if X is $$\mathcal{G}$$-measurable, and $$E[X|\mathcal{G}]=X$$, if X is independent to $$\mathcal{G}$$. Thus, we can conclude that if the random variable is independent of $$\mathcal{G}$$, then X is not $$\mathcal{G}$$-measurable.