# Definition of lim sup when taking the sequence of inf

Just want to gain clarification on lim sup as used in the Borel-Cantelli and Fatou's lemma.

Is the following statement true by definition?

$$\lim\limits_{n\rightarrow\infty}\inf f_n=\lim\limits_{n\rightarrow \infty}\inf\limits_{m\geq n} f_m$$

My question is that the usual definition of $$\lim\inf$$ is $$\sup\inf$$, so I am trying to understand how does taking the sup of the infs of the tail sequence of sets or measurable functions related to actually taking the limit of the infs of the tails sequence of sets or measurable functions?

• Yes, this is the definition. Similarly, $\lim_{n\to\infty} \sup f_n=\lim_{n\to\infty}\sup_{m\geq n}f_n$ – zugzug Feb 17 '20 at 15:43
• I think that should be $f_m$ on the right-hand side, no? – saulspatz Feb 17 '20 at 15:44
• I made clarification. Thanks. – Frank Swanton Feb 17 '20 at 15:46
• @FrankSwanton The key insight is that the sequence of tail infima is nondecreasing. Therefore, the limit and the supremum are the same. – David Kraemer Feb 17 '20 at 15:51
• @DavidKraemer Excellent, David. Thanks. Can you provide an example where they differ? A simple example to have in my pocket. – Frank Swanton Feb 17 '20 at 16:01

Consider the sequence $$(x_n)_{n=1}^{\infty}$$ defined $$x_n = \begin{cases} 1 + n^{-1} & \text{if n is even,} \\ -1 - n^{-1} & \text{if n is odd,} \end{cases} = (-1)^{n}(1+n^{-1}).$$ Some casual observations about this sequence include (a) the odd part and the even part are identical, just mirrored, (b) the sequence has no limit, (c) the sequence has two unique subsequence limits: $$-1$$ and $$1$$. The point of this sequence is to illustrate the definitions of the limits superior and inferior. I'll focus on the $$\limsup$$ and you can think through the details for the $$\liminf$$.
Using the definition $$\limsup_{n \to \infty} x_n = \inf_{n \in \mathbb{N}} \sup_{m \geq n} x_m$$, let's figure out what the "sup sequence" looks like. In particular, for a given $$n \in \mathbb{N}$$, we have $$\sup_{m \geq n} x_m = \sup_{m \geq n} (-1)^{m} (1 + n^{-1}) = 1+n^{-1}.$$ This follows from two observations. First, we can discard the lower subsequence because it is always bounded above by the upper sequence. Second, the upper subsequence is decreasing, which means its largest value is its first value. Whence $$1 + n^{-1}$$. It therefore follows that \begin{align} \limsup_{n \to \infty} x_n = \inf_{n \in \mathbb{N}} 1 + n^{-1} = 1. \end{align} Now, if we had instead defined $$\limsup_{n \to \infty} x_n$$ as $$\lim_{n \to \infty} \inf_{m \geq n} x_m$$, we would follow the same steps and compute \begin{align} \limsup_{n \to \infty} x_n = \lim_{n \to \infty} 1 + n^{-1} = 1. \end{align} The point is that the sequence of suprema are necessarily nonincreasing. In that case, the infimum and limit coincide (this is the monotone convergence theorem for sequences).
Aside. Why is the sequence of suprema nonincreasing? The simple fact that $$A \subset B$$ implies $$\sup A \leq \sup B$$ allows us to say that, for example, $$\sup \{x_m : m \geq 2\} \leq \sup \{x_m : m \geq 1\}$$ and so forth.
Finally, there is the alternate characterization of the limit superior: $$\limsup_{n \to \infty} x_n = \sup \{\bar{x} \in \mathbb{R} : \exists x_{n_k} \to \bar{x} \}.$$ The set on the right hand side is also called the "cluster points" of $$x_n$$, composed of the limits of subsequences of $$x_n$$. It is a worthwhile exercise to prove that this characterization works. In the meantime, we can verify it for our toy sequence by observing that $$\{-1,1\}$$ are the cluster points of $$x_n$$, and $$\limsup_{n \to \infty} x_n = 1$$.