Let $(\Omega,\mathscr{F},\mu)$ be a probability space and $\mathscr{G}\subseteq \mathscr{F}$ a $\sigma$-algebra. I have seen it referred to many times that $\mathscr{G}$ is the "information" which is available to us. I think I kinda understand it, but I am not satisfied in my own understanding of it.

Let us say that $A$ is an event and $\mathscr{G} = \{\emptyset, A,A^c,\Omega\}$. Let $\omega$ be a sample point of the experiment. We do not know which of the events in $\mathscr{G}$ contain $\omega$. Now $\Omega$ certaintly contains $\omega$, but we already knew that, so we do not gain any insight. However, either $A$ or $A^c$ will contain $\omega$. If we somehow knew that $A$ contains $\omega$, then that gives us additional insight.

1) Does anyone have a better way of thinking of $\mathscr{G}$ as our "information"?

2) If $\mathscr{G}'\supseteq \mathscr{G}$, then how do we think of $\mathscr{G}'$ as having "more information"? Obviously, it is a larger $\sigma$-algebra, and it has more events, but ignoring set theory, what should one's intuition be for $\mathscr{G}'$?

Follow up question.

3) Let $\xi:\Omega\to\mathbb{R}$ be a $\mathscr{G}$-measurable. I have seen people refer to $\xi$ as "a random variable whose information is known from $\mathscr{G}$", or something along those lines. What is the motivation for this?

  • $\begingroup$ I don't really understand your first sentence. What is G, just any old sub-$\sigma$-algebra of F? I see that F is the $\sigma$-algebra needed to define the probability space, but what's G doing here? $\endgroup$ Aug 21, 2016 at 6:00
  • $\begingroup$ @Kodiologist When you condition on a $\sigma$-algebra G, G is often referred to as "information that you know". $\endgroup$ Aug 21, 2016 at 6:05
  • $\begingroup$ Huh? But you can't condition on G. It's not an event. $\endgroup$ Aug 21, 2016 at 6:08
  • 2
    $\begingroup$ @Kodiologist see, e.g., en.wikipedia.org/wiki/… ("see" meaning "see that stuff conditional on sigma-algebras is indeed talked about", not "read and learn" for which the Wikipedia article is probably not the best place). $\endgroup$ Aug 21, 2016 at 6:48
  • $\begingroup$ @Kodiologist You should delete your comments. They are not about the nature of my question, but your question about conditioning on $\sigma$-algebras. Which is a separate question in itself. $\endgroup$ Aug 21, 2016 at 7:33

1 Answer 1


$\mathscr{G}$ is our information in the sense that for all $A \in \mathscr{G}$, we know whether $\omega \in A$.

Let us use the Tickets in a box metaphor, extended to handle $\sigma$-algebras so that the ticket mentions for all $A\in \mathscr{F}$ whether the outcome represented by the ticket belongs to $A$. Now, say that someone else picks the ticket and we don't see it. For any $A \in \mathscr{G}$ we may ask whether the ticket says that the outcome is in $A$ and the person holding the ticket tells us. However, if we ask about some $A \in \mathscr{F} \setminus \mathscr{G}$, we hear "Sorry, you don't know that".

Larger $\sigma$-algebra is more information

This also explains why moving to $\mathscr{G}' \supset \mathscr{G}$ means gaining new information -- now we still get answers to $[X \in A?]$-questions about any$A \in \mathscr{G}$ and additionally to some new questions -- those where $A \in \mathscr{G'} \setminus \mathscr{G}$.

Random variables

So, the tickets also contains the values of random variables. If the random variable $X$ is $\mathscr{G}$-measurable, we get answers to all our questions about its value, such as $[$is $X$ equal to $3]$, since by $\mathscr{G}$-measurability of $X$, $\{\omega \mid X(\omega)=3\}\in\mathscr{G}$. Or, to handle the delicacies of the uncountable case, we may also ask $[$Is $X$ in the set $B]$? (Since for any particular value we think about, the probability of hearing "yes" may be $0$ and that would be boring). So, in this sense we have all information about the realization of the random variable, if our information is $\mathscr{G}$ and the RV is $\mathscr{G}$-measurable.

Caveat: the definition of measurability of random variables restricts the sets $B$ we may ask about. $[$Is $X(\omega) \in B]$ is answered if $B$ is a measurable set in the value space of the random variable (usually Borel $\sigma$-algebra is assumed with $\mathbf{R}$ without mentioning). So, in the uncountable (nondiscrete $X$) case, don't ask whether $X$ is in the Vitali set or the oracle holding the ticket shall be mad.


I did not cite any reference in the answer but I consulted

  • J. Jacod and P.E. Protter. Probability essentials (2nd edition), Springer, 2004

about the definition of measurability of random variables. (And have learned these things from the same book previously, if I recall correctly).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.