# Random Variable Measurability w.r.t. Sigma Fields

1. Sep 14, 2011

### empyreandance

Hello everyone,

I'm having a little trouble with a probability problem with three parts; I think I'm having trouble wrapping my head around just what's going on here. If anyone could give me a starting point, I'd appreciate it.

Here's the problem (Billingsley 5.1) (X a random variable)

a. Show that X is measurable w.r.t. the sigma field J iff sigma(X) is a subset of J. Show that X is a measurable w.r.t. sigma(Y) iff sigma(x) is a subset of sigma(Y)

b. Show that if J = {empty set, omega}, then X is measurable w.r.t. J iff X is constant.

c. Suppose that P(A) is 0 or 1 for every A in J. This holds, for example, if J is the tail field of an independent sequence, or if J consists of the countable and cocountable sets on the unit interval with Lebesgue measure. Show that if X is measurable w.r.t. J, then P[X=c] = 1 for some constant c.

Thanks for any and all help!

Best regards

2. Sep 14, 2011

### micromass

Staff Emeritus
Did you see topology already?? The techniques used here ressembles the techniques in topology.

Anyway, for (a), you need to prove that if X is measurable, then $\sigma(X)$ is a subset of J.

So, firstly, how is $\sigma(X)$ defined?

3. Sep 14, 2011

### empyreandance

Hello,

I thought there might be some sort of topological argument, but the book is very analysis-oriented, so I was trying to stick to that line of thinking. σ(X) is defined as the smallest sigma field that X is measurable w.r.t., i.e. the intersection of all such fields.

4. Sep 14, 2011

### micromass

Staff Emeritus
OK, so $\sigma(X)$ is the smallest sigma field such that X is measurable. Doesn't that make it easy to show that $\sigma(X)\subseteq J$?? X is measurable w.r.t. J after all...

5. Sep 14, 2011

### empyreandance

Ah yes, of course it does. I'm not sure why I missed that. Thanks. As for the other parts, any suggestions?

6. Sep 14, 2011

### micromass

Staff Emeritus
For the reverse, you know that $\sigma(X)\subseteq J$and that $\sigma(X)$ makes X measurable. You need to show that J makes X measurable. So by going to a finer sigma field, you preserve measurability. That shouldn't be too difficult?

7. Sep 14, 2011

### empyreandance

No, it's not bad at all. It's more parts b and c that I remain a bit lost on

8. Sep 14, 2011

### micromass

Staff Emeritus
OK take X to be measurable wrt the trivial sigma field. We wish to prove that X is constant.

What is $\{X=c\}$ for each c in $\mathbb{R}$?? A measurable set w.r.t. the trivial sigma field,right?? What can you deduce?

9. Sep 14, 2011

### empyreandance

Yes, definitely measurable. Oh! If X is not constant, then the inverse image maps to a set strictly smaller than the space, or am I completely confused now? My apologies, for some reason I'm having particular difficulty on this one.

10. Sep 14, 2011

### micromass

Staff Emeritus
OK, $\{X=c\}$ is a measurable set wrt to the trivial sigma-algebra. But the trivial sigma-algebra only has two measurable sets: $\emptyset$ and $\Omega$.

So what are the only possibilities for $\{X=c\}$??

11. Sep 14, 2011

### empyreandance

Ah, either the empty set or the entire space. So, for {X=C} to be measurable it either has to be the empty set or omega.

12. Sep 14, 2011

### micromass

Staff Emeritus
Yes! And what does it mean that $\{X=c\}=\Omega$??

13. Sep 14, 2011

### empyreandance

It means the set of x in omega s.t. X(x) = c is the entire space, correct?

14. Sep 14, 2011

### micromass

Staff Emeritus
So, if $\{X=c\}=\Omega$, then the function is constant, right??

What you have proven is that either $\{X=c\}$ is empty or is omega. If there is a c such that $\{X=c\}$, then you're done.

Now prove that there must exist such a c. Hint: what is $\{X\in \mathbb{R}\}$??

15. Sep 15, 2011

### empyreandance

Yes, the function is assuredly constant. X in R is the set of omega such that X(w) is in R. However, X is a random variable, so it maps from the the space to R. So, I feel the connections starting to coalesce in my head... I think now, Since J is the empty set or the whole space and we've shown that either {X=c} is empty or is omega, then since X in R is the set of omega such that X(w) is in R, which should be the entire space. Then, since if a X(y) = d, whereas X(x) = c elsewhere, both these sets would map back to sets strictly smaller than the space, but strictly larger than the empty set by the way they're defined, yes?

16. Sep 15, 2011

### micromass

Staff Emeritus
Seems alright!!

17. Sep 15, 2011

### empyreandance

Awesome, I can't thank you enough for both your help and patience!