# Proving the R.V. condition

• Oct 11th 2009, 10:59 AM
mliang87
Proving the R.V. condition
Hello all, I have a problem that I'm not too clear about.

It is as follows:

Let X be a random variable. Show that for every x E R,
B(x) = { w E Omega: X(w) < x} E F

*Where E = element of; Omega = collection of objects, the sample space; w = a member of Omega, elementary or outcome; and F = a field

My solution begins with setting,
B_n(x) = {x E Omega: X(w) <= x - 1/n}

Then I am not too sure how to go about proving it...

I was thinking along the lines of showing equivalence between B(x) and B_n(x), since B_n(x) is already defined to be in the field...

• Oct 11th 2009, 11:06 AM
mliang87
posted solution
Actually, here is the solution that was written on the board, but I wasn't able to follow it at some parts.

Let x E R. Let n >= 1 be an integer.

Set B_n(x) = {xE Omega: X(w) <= x - 1/n} E F

Then B_n(x) is a subset of B_n for all n >=1.
So, the Union where n=1 to infinity B_n(x) is a subset of B(x). Show B(x) = Union from n=1 to infinity of B_n(x).

Claim: B(x) is s subset of Union from n=1 to infinity of B_n(x). Else there exists w E B(x) s.t. w E/ Union from n=1 to infinity of B_n(x)

So for every n >=1, [w E Union from n=1 to infinity of B_n(x)]^C or X(w) > X - 1/n.

(***) Letting n --> infinity, we see

X(w) >= x

Consequently, if w E B(x)
X(w) < x and X(w) >= x

This is impossible and thereforefore B(x) = Union from n=1 to infinity of B_n(x) E F as B_n(x) E F for each n.

(***) This is where I get completely lost.

Also, ^C represents the complement