Can a Bounded Random Variable Be Found for Almost Equal Random Variables?

[student12s]

I've been trying to solve the following question: Let X be a random variable s.t. Pr[|X|<+\infty]=1. Then for every epsilon>0 there exists a bounded random variable Y such that P[X\neq Y]<epsilon.

The ideia here would be to find a set of epsilon measure so Y would be different than X in that set. However, it is not clear even that such a measurable set exists...

Any help?

 

[mXSCNT]

To show there is some interval (-a, a) such that $$\int_{-a}^a f(x) dx > 1-\epsilon$$, apply the definition of an improper integral to $$\int_{-\infty}^\infty f(x) dx = 1$$. Have Y=X over (-a,a).

 

[g_edgar]

For a large positive number $\lambda$, define $Y = X$ on the set $\{|X| \le \lambda\}$ and $Y=0$ on the remaining part of the sample space. Now all you have to do is choose $\lambda$ large enough to get the conclusion you want.

 

[student12s]

It would remain to show that such $\lambda$ exists. We can definte the sets A_n=\{\omega\in \Omega: |X(\omega)|<n\}. Since A_n is measurable, as X is a random variable, and A_n \uparrow \{\omega: |X|<+\infty}, we have that \lim P(A_n) = P(\{\omega: |X|<+\infty}. Therefore, for all \epsilon, there is $m$ large enough so that P(A_n)>1-\epsilon. Then we can define Y=X in A_m and Y=0 otherwise.

Thanks!