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If $X$ is a random variable satisfying $P[|X| \lt \infty]=1$, then show that for any $\epsilon >0$, there exists a bounded random variable $Y$ such that $P[X \neq Y] \lt \epsilon$.

Note: A random variable $Y$ is bounded if for all $\omega$, $|Y(\omega)| \le K$ for some $K$ independent of $\omega$.

I could not understand this question and solve it.

Thanks and regards.

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    The question is asking you to find a random variable $Y$ that is (a) bounded, as you describe, and (b) equal to $X$ on the whole space excepting a set of arbitrary small measure. The condition on $X$ is that it is almost surely finite. So, think about what that says about the measure of the set on which $X$ takes "large" values, i.e., |X| > B for some large $B$.2011-11-18

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I am just making cardinal's point precise. Since $P(|X|<\infty)=1$, $P(\bigcap_n\{|X|>n\})=0$, and hence $\lim_{n\to \infty}P(|X|>n)=0$.

Now, let $Y_n=X\cdot 1_{\{|X|\le n\}}$. Then $Y_n$ are all bounded and $\{X\neq Y_n\}=\{|X|>n\}$.

Hence , for given $\epsilon>0$, there exists $Y_n$ such that $P(X\neq Y_n)<\epsilon$.

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    It means the indicator $I(|X|\le n)$2017-08-16