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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? –  Mary Dec 12 '10 at 18:46
    
You are talking about the limsup of a sequence of sets. How is that defined? –  Tsuyoshi Ito Dec 12 '10 at 19:00
    
en.wikipedia.org/wiki/… –  Mary Dec 12 '10 at 19:03
    
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. –  Raskolnikov Dec 12 '10 at 19:04
    
@ 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 positive –  Mary Dec 12 '10 at 19:10
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2 Answers 2

up vote 4 down vote accepted

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? –  Mary Dec 12 '10 at 19:21
    
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. –  Raskolnikov Dec 12 '10 at 19:24
    
@Mary: That's probably true, but I'll check this. –  Shai Covo Dec 12 '10 at 19:28
    
@ 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? –  Mary Dec 12 '10 at 19:29
    
Just if the sets are the same. –  Raskolnikov Dec 12 '10 at 19:42
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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? –  Mary Dec 12 '10 at 20:42
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