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Let $(\Omega, \mathcal{F}, P)$ be a probability space, where $\Omega = [0,1]$, $\mathcal{F}=\mathcal{B}(\Omega)$ is a Borel $\sigma$-algebra and $P$ is a Lebesgue measure. Given a sequence of random variables defined as follows: $$X_n(\omega) = n \cdot \mathbb{1}_{\left[0,\frac{1}{n}\right]}(\omega)$$ I need to check the different types of convergence of $\{X_n\}$, that is in probability, almost sure, in mean and in distribution. At first, I thought that this sequence is divergent, since I don't understand what would be the value of $X(0)$ (if $X$ is a limit), though the answer in my source is quite different.

Thanks to all!

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In distribution it converges to $0$, since the probability that $\omega$ belongs to $(1/n,1)$ tends to $1$, thus the indicator function will 'always' be $0$ in the limit.

Please note that the value in a certain point, for example $0$, does not influence the distribution of$X$ (since it is a continuous random variable).

This is not a complete answer, because I don't think that will be the most useful. However, if you have any question, please let me know!

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I wouldn't worry about the divergence of $X_n(\omega)$ for $\omega = 0$. The singleton set $\{ \omega \in \Omega : \omega = 0 \}$ has measure zero! What's important is that $\lim_{n \to \infty} X_n(\omega) = 0$ for $\omega \in \Omega \setminus \{ 0 \}$ (as you can easily verify). Hence $X_n$ tends to zero almost surely, i.e. $X_n$ tends to zero everywhere except on a set of measure zero.

Having verified that $X_n \to 0$ almost surely, it follows immediately that $X_n \to 0$ in probability, and in distribution. (Or you can check explicitly...)

It only remains to check if $X_n$ converges to anything in mean. If $X_n$ were to converge to anything in mean, then it would have to converge to $0$ (because convergence in mean implies convergence to the same limit in probability). But since $\int_\Omega |X_n - 0| = 1$ for each $n \in \mathbb N$, it's clear that $X_n$ does not converge to $0$ in mean. (Alternatively, if you prefer a different approach, you can show that $X_n$ is not Cauchy in $L^1(\Omega, P)$ ...)

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  • $\begingroup$ Thanks, exactly this thing was unclear. Since in all the definitions above we first fix a limit $X$ and then check if it satisfies a properties, I didn't know, how to deal with it if it's divergent $\endgroup$ – Denisof Jun 22 '18 at 22:15

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