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I have a question about whether the following two sets on the two sides are the same:

$$ \lim_{a \rightarrow \infty} \, \limsup_{n \rightarrow \infty} \, \{ x \in S \, | \, f_n(x) > a \} = \left\{ x \in S \, \Bigm| \, \lim_{n \rightarrow \infty} f_n(x) =\infty \right\}\quad ? $$

where $\{f_n\}$ is a sequence of real-valued functions defined on a set $S$. How can you explain it?

Thanks in advance!

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    Here is what I understand, on the left hand side, first take the limsup of a sequence of sets and the result is still a set, which is parameterized by a, and then let a goes to infinity. So the lhs is a set, isn't it?2010-12-12
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    You are talking about the limsup of a sequence of sets. How is that defined?2010-12-12
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    http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Sequences_of_sets2010-12-12
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    Try with a sequence of functions that has an alternating component that gets larger in absolute value with $n$ but changes sign to see if this still holds. It shouldn't if I read the definition of limsup for sets correctly.2010-12-12
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    @ Raskolnikov: can you explain more? How about if {f_n} are all positive functions? I tend to think the two sets are the same, and don't see it make differences whether {f_n} are positive2010-12-12
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    Sorry, you didn't mention that all the f should be positive. But even then, I still think it is not true. Make a sequence that is defined as follows: $f_n(x)=n/(1+(nx)^2)$ for $n$ even and $f_n(x)=1$ for $n$ odd.2010-12-12
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    @ Raskolnikov:Thanks! I see. Is it true that the left set always a super set of the right set? Note that I have changed the domain of f_n from R to any set S earlier.2010-12-12

2 Answers 2

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Consider a sequence of functions $f_n:\mathbb{R} \to \mathbb{R}$ defined by $f_n {(x)} = n$ if $x=0$ and $n$ is even, and $f_n {(x)} = 0$ otherwise. The right-hand side is obviously $\emptyset$, while the left-hand should be $\lbrace 0 \rbrace$.

EDIT: Answering the additional questions.

Suppose that $f_n{(x)}$ is an increasing sequence for each $x$. If $x$ belongs to the left set, then it must belong to the set $\lim \sup _{n \to \infty } \{ x \in S|f_n (x) > a\}$ for any $a > 0$ fixed. This means that for any $a>0$, there are infinitely many $n$ such that $f_n (x) > a$. Since $f_n (x)$ in increasing, it must converge to a positive number, or diverge to $\infty$. But $a>0$ is arbitrary, hence $f_n (x)$ must diverge to infinity. That is, $x$ belongs to the right set. Since the right set is contained in the left one, we conclude that both sets are equal.

EDIT: The fact that the right set is contained in the left one, is proved as follows. Suppose that $x$ belongs to the right set, and let $a>0$ be arbitrary but fixed. Then, for all sufficiently large $n$, $f_n {(x)} > a$. In particular, $x \in \lim \sup _{n \to \infty } \{ x \in S|f_n (x) > a\}$. Since this is true for any $a > 0$, $x$ belongs to the left set.

EDIT: The following point should be stressed. Denote by $E_a$ the set $\lim \sup _{n \to \infty } \{ x \in S|f_n (x) > a\}$. If $x \in E_a$, then $x \in E_{a'}$ for any $a' < a$. It follows that $\lim \sup _{a \to \infty} E_a = \lim \inf _{a \to \infty} E_a$; hence, by definition, the limit $\lim _{a \to \infty} E_a$ exists, and is equal to $\lim \sup _{a \to \infty} E_a = \lim \inf _{a \to \infty} E_a$. So, the left set in the question is indeed properly defined, and $x \in \lim _{a \to \infty} E_a$ means, in our case, that $x \in E_a$ for every $a$. Finally, note that always the $\lim \inf$ is a subset of the $\lim \sup$ (in analogy with the case of sequences of real numbers, where $\leq$ plays the role of $\subseteq$).

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    @Shai:Thanks! Is it true that the left set always a super set of the right set?2010-12-12
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    Makes me wonder if replacing the lim in the right hand side with a limsup makes it true? I still think it's not, but I'm not sure.2010-12-12
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    @Mary: That's probably true, but I'll check this.2010-12-12
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    @ Raskolnikov: are you talking about if it is true that the two sets are same, or if the left one is a superset of the right one?2010-12-12
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    Just if the sets are the same.2010-12-12
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    What if furthermore for each x in S, $\{ f_n(x) \}$ is an increasing sequence? Are the two sets are the same now?2010-12-12
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    A strictly increasing sequence? If not, I got another counter example I think.2010-12-12
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    Yes, strictly increasing.2010-12-12
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    OK, got one the same.2010-12-12
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    @ Raskolnikov: Thanks! How do you show they are same when {f_n(x)} is an increasing sequence for each x?2010-12-12
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    They are not always the same, see my counterexample.2010-12-12
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    @Shai : It need not always be the same $x$ that is bigger than $a$ at each step. You can only prove inclusion of the right hand into the left hand.2010-12-12
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    General remark: There might be different definitions, leading to a confusion. It may be useful to consider this thread http://mathoverflow.net/questions/12462/limsup-and-liminf-for-a-sequence-of-sets2010-12-12
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    Yes, I see there are different definitions of limsup for sets. Mary should specify which one she needs.2010-12-12
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    @Raskolnikov: What definitions were you using? I thought the definition for lim of a sequence of sets is when limsup and liminf of the sequence agree. The definitions for limsup and liminf of a sequence of sets are specified in the first sentence of one reply to the question linked by Shai Covo http://mathoverflow.net/questions/12462/limsup-and-liminf-for-a-sequence-of-sets/12497#12497 and my previous comment to Tsuyoshi Ito.2010-12-12
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    @Shai : Thanks! Is it true that the right set is contained in the left one? How to show it?2010-12-12
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    @Mary: Yes, this is shown in my edited answer.2010-12-12
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    I used the definition on the wiki page which says there should be a subsequence of sets and within that a sequence of points converging to a point $x$ for $x$ to be in the limsup. I guess it is a looser definition than the one in your link, and thus the example I constructed will probably not work.2010-12-12
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    I use the general definition http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#General_set_convergence.2010-12-12
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    That's the one I used, but it is different from Shai's definition. In Shai's definition, any element $x$ contained in the limsup should be contained in each element of the series. This is not true in the general set convergence definition.2010-12-12
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    @Raskolnikov: I think Shai uses the same definition as ours. I don't see he is assuming "any element x contained in the limsup should be contained in each element of the series". He said "for any a>0, there are infinitely many n such that fn(x)>a."2010-12-12
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    See the MO page you referred to, it is stated explicitly there.2010-12-12
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    @Mary: I don't use the same definition. Apparently, the general definition in Wikipedia is not the common definition -- why do you think you should use it?2010-12-12
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    @ Shai: My mistake. I have been using the definition in the first sentence to the reply in MO http://mathoverflow.net/questions/12462/limsup-and-liminf-for-a-sequence-of-sets/12497#12497, which I think is same as the Special_case:_discrete_metric in wikipedia http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#Special_case:_discrete_metric. Are you using the same definition as mine?2010-12-12
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    @Mary: Yes, we are now using the same definition (which is also the common definition).2010-12-12
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For question number two, construct the following sequence of $f_n(x)$

$$ f_n(x) = \left\{\begin{array}{ll} k+\epsilon_n, & \mbox{for $x=1/k$ with $k=1,\ldots,n$,} \\ \epsilon_n, & \mbox{otherwise.}
\end{array}\right.$$

With the sequence of $\epsilon_n=1-\frac{1}{2^n}$.

If I did not make any mistake, the right-hand side should be empty, while the left-hand side contains $0$. (This example was constructed with the general set convergence in mind.)

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    Thanks! your example is a wonder. How to understand the left-hand side contains 0?2010-12-12