Let ${\left\{{f}_{k}\right\}}_{k=1}^{\mathrm{\infty}}$ be a sequence of $\mathcal{S}$-measurable real-valued functions on a measure space $(M,\mathcal{S})$. Then the functions

$\underset{k\to \mathrm{\infty}}{lim\u2006sup}{f}_{k}(x)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{k\to \mathrm{\infty}}{lim\u2006inf}{f}_{k}(x)$

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 ${\left\{{a}_{k}\right\}}_{k=1}^{\mathrm{\infty}}$ of real numbers the following is true:

$\underset{k\to \mathrm{\infty}}{lim\u2006sup}{a}_{k}=\underset{k\to \mathrm{\infty}}{lim}\underset{m\to \mathrm{\infty}}{lim}max\{{a}_{k},{a}_{k+1},\dots ,{a}_{m}\}$

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 $lim\u2006sup,lim\u2006inf$ like that. I know that for a set of limit points $H$ we can say that $lim\u2006supH=maxH$ and vice versa. I do not even see how this property can be useful for our proof.