3
$\begingroup$

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!

  • 0
    @ 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

4

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 $\subseteq$).

  • 0
    @Mary: Yes, we are $n$ow using the same definition (which is also the common definition).2010-12-12
1

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.)

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