Let $\left\{f_{k}\right\}_{k=1}^{\infty}$ be a sequence of $\mathcal{S}$-measurable real-valued functions on a measure space $(M, \mathcal{S})$. Then the functions $$ \limsup _{k \rightarrow \infty} f_{k}(x) \quad \text { and } \quad \liminf _{k \rightarrow \infty} f_{k}(x) $$ are also $\mathcal{S}$-measurable.
I already have proven this statement and my question is about something else. In the script it says that the reader may use the hint:
For a sequence $\left\{a_{k}\right\}_{k=1}^{\infty}$ of real numbers the following is true: $$ \limsup _{k \rightarrow \infty} a_{k}=\lim _{k \rightarrow \infty} \lim _{m \rightarrow \infty} \max \left\{a_{k}, a_{k+1}, \ldots, a_{m}\right\} $$
Now, I have not used this in my proof, but I can assure you, that my proof is airtight nonetheless. Anyway, can someone point out why this holds? I have not seen this equality before and did not know that one can present $\limsup, \liminf$ like that. I know that for a set of limit points $H$ we can say that $\limsup H = \max H$ and vice versa. I do not even see how this property can be useful for our proof.
By definition $$\limsup_{k\rightarrow \infty} a_k = \lim_{k\rightarrow \infty} (\sup_{m\geq k} a_m),$$ so to prove the claim, it suffices to prove for any $k\in \mathbb{N}$ that $$\sup_{m\geq k}a_m = \lim_{m \rightarrow \infty} \max(a_k, a_{k+1},\dots,a_{k+m}).$$
The reason that this representation is useful, is that $m \mapsto \max(a_k,\dots,a_{k+m})$ is increasing, and $k \mapsto \sup_{m\geq k} a_m$ is decreasing. So to prove that $\limsup_{k\rightarrow \infty} f_k$ is measurable, it suffices to show that the limit of monotone sequences of functions are measurable.