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Here is the question: Suppose $X$ is a $\Bbb R$-valued random variable. Show that for all $\varepsilon > 0$, there exists a bounded random variable $Y$ such that $P (X \neq Y ) <\varepsilon$.

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up vote 2 down vote accepted

Try $Y=X\,\mathbf 1_{|X|\leqslant x}$ for $x$ large enough. Then $\mathbb P(Y\ne X)=\mathbb P(|X|\gt x)$. Furthermore, $\mathbb P(|X|\gt x)\to0$ when $x\to+\infty$, hence for $x$ large enough, $Y$ solves the question.

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I presume that $P(X=Y)$ in the question is a typo and should read $P(X\ne Y)$. – Did Oct 12 '12 at 22:37
What about the part {X>x}. Could you explain a little bit more? – 65900931 Oct 13 '12 at 1:10
Sorry but I do not understand the question in your comment. If explain a little bit more means provide a full solution one can hand as it is to one's teacher, the answer is no. Otherwise, please explain the part you do not understand. – Did Oct 13 '12 at 8:25
It's obviously Y can cover $X\,1_{\left|X\right|\leq x}$ for arbitrary x. But how do you deal with the part $X\,1_{\left|X\right|>x}$? How can it be less than any positive number when you fix a x? Thanks! – 65900931 Oct 13 '12 at 17:16
Cover? The part? Fix $x$? See the expanded version of my answer. – Did Oct 13 '12 at 18:01

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