I am asked to show :
Let $X$ be a real-valued random variable on $(\Omega, F , P)$ and define
$X_n(\omega) = nX(\omega)$ if $n<X(\omega)\le n+1$ and $0$ if else. Prove that $X_n \to 0$ in probability.
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2$\begingroup$ (+1) Welcome to Stats.SE. You might be interested to know that you can use math typesetting via Mathjax. More information: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Sycorax♦ Jan 14 at 19:41
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1$\begingroup$ Are you sure you stated the problem as intended? By writing "$(\Omega,F,P),$" you suggest $\Omega$ is an abstract set, but in writing "if $n\lt w\le n+1$" you imply the positive real numbers are in $\Omega.$ $\endgroup$ – whuber♦ Jan 14 at 19:47
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1$\begingroup$ That's not the issue: the issue is that the argument of $X,$ which you call "w," must be a real value for this problem statement to make sense. At the very least, your book/exam/professor is making assumptions or restrictions that you haven't put in evidence here. $\endgroup$ – whuber♦ Jan 14 at 19:53
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1$\begingroup$ I don't see any use of indicator functions in your question. You might benefit from drawing graphs of some of the $X_n$ and contemplating what happens to the chances that their values exceed zero. What I wonder is whether the clause "$n\lt w\le n+1$" should instead read "$n\lt X(w)\le n+1.$" That would make sense for any abstract probability space; it would imply the same conclusion; but would require a different solution. $\endgroup$ – whuber♦ Jan 14 at 20:09
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2$\begingroup$ Please, then, edit your post to reflect that and add the self-study tag after reading information about it. $\endgroup$ – whuber♦ Jan 14 at 21:18
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I think the easiest way is to show that $X_n(\omega)\stackrel{as}{\to}0$ and note that almost sure convergence implies convergence in probability.
$X_n(\omega)=0$ for all $n>X(\omega)$, so for every fixed $\omega$, $X_n(\omega)\to 0$.
Therefore $X_n(\omega)\stackrel{as}{\to}0$, and it is a standard fact that this implies $X_n(\omega)\stackrel{p}{\to}0$
