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Suppose $A_n$ is a sequence of sets such that $\lim_{n\to\infty}A_n = A$ where we make no assumption on the $A_n$ being increasing or decreasing. Is $\lim_{n\to \infty}\chi_{A_n} = \chi_{A}$?

For example consider a sequence in the reals $x_n$ that has limit $x$ then define $A_n = [0,x_n]$. Is it true that $\lim_{n\to\infty} \chi_{[0,x_n]} = \chi_{[0,x]}$?

If the example given does not hold what are some conditions such that it does?

Thanks :)

When I say $\chi_S$ I mean the indicator of set $S$, maybe you have seen it as $\mathbb{1}_S$ :)

OskarSzarowicz
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    If the $A_n$ are not assumed to be nested (non-increasing or non-decreasing), how is $\lim A_n$ defined? – coffeemath Apr 25 '21 at 13:17
  • Could you clarify what each of the limits means? What does it mean for a sequence of sets $A_n$ to converge to $A$? What kind of convergence are you looking at? Pointwise? Uniform? In measure? – Theo Bendit Apr 25 '21 at 13:17
  • Some information on general limits of sets can be found here – Snoop Apr 25 '21 at 13:19
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    Often (at least, in the limsup-and-liminf-of-set jargon that accompanies some probabilistic topic) $\lim_{n\to\infty}A_n$ is defined as the set $U$ such that $\chi_U(x)=\lim_{n\to\infty}\chi_{A_n}(x)$ for all $x$. –  Apr 25 '21 at 13:31
  • @TheoBendit Truthfully I am only interested in the example I gave about a sequence of reals. But I gave a more general concept. I see now defining the limit of a set is foolish! Do you know if the example I gave is true. Thanks :) – OskarSzarowicz Apr 25 '21 at 13:33
  • @coffeemath I see now there is no such clear thing as the limit of a set. All i really cared about what the example I gave but I made a more general question. I will edit it :) – OskarSzarowicz Apr 25 '21 at 13:34
  • @Gae.S. I think you might have made a little typo somewhere. I split the cases into $t \leq x$ and $t > x$ Thank you for contributing :) – OskarSzarowicz Apr 25 '21 at 13:46
  • @coffeemath: $\lim A_n$ is defined as $\liminf A_n=\limsup A_n$, in case the two happen to be equal. See https://math.stackexchange.com/questions/107931/lim-sup-and-lim-inf-of-sequence-of-sets – Hans Lundmark Apr 25 '21 at 14:51
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    @BearCakes100 Yes, the previous comment was wrong. The correct version is $$\limsup_{n\to\infty}\chi_{[0,x_n]}(t)=\begin{cases}\chi_{[0,x]}(t)&\text{if }x_m\ge x\text{ frequently}\ \chi_{[0,x)}(t)&\text{if }x_m<x\text{ eventually}\end{cases}\ \liminf_{n\to\infty}\chi_{[0,x_n]}(t)=\begin{cases}\chi_{[0,x]}(t)&\text{if }x_m\ge x\text{ eventually}\ \chi_{[0,x)}(t)&\text{if }x_m<x\text{ frequently}\end{cases}$$ hence the sequence converges at $x$ if and only if either $x_m\ge x$ eventually or $x_m< x$ eventually, in which case it converges everywhere. –  Apr 25 '21 at 16:29

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I found an answer to the example I gave and that is yes.

Assume $x_n \to x$ then for any $t \in [0,x]$ i.e $0 \leq t \leq x$ we have that $\forall \epsilon >0 \exists N \in \mathbb{N}$ such that $n>N \implies x -\epsilon < x_n $ (and something on the RHS that we dont care about yet)

Then taking $\epsilon = x-t $ we have $\forall n > N$ $ t < x_n$ and so $t\in[0,x_n]$ and so $\chi_{[0,x_n]}(t) = 1 = \chi_{[0,x]}$ Hence the limit holds true for all $t \in [0,x]$

Similarly for $t > x$ we have $\exists N \in \mathbb{N}$ such that (taking $\epsilon = t - x$) $ n > N \implies x_n < t$ i.e $t\not\in[0,x_n]$ for all $n>N$ and so $chi_{[0,x_n]}(t) = 0 = \chi_{[0,x]}(t)$ so again the limit holds true.

The final case to analyse is $t < 0 $ however this is trivial

Hence we are done.

OskarSzarowicz
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