I have come across suspicious use of liminf a few times, while reading the book. So I decided to post one short case because it seems to me I am probably understanding it wrong. For me limsup does much better job. But maybe I am totally wrong, in which case I would be glad if you let me know and put here some explanation why you think so. First of all I cite the book.
3.17 Lema $L^\infty$ is complete.
Proof: Let $(f_k)_{k\in \mathbb{N}}$ be a Cauchy sequence in $L^\infty(\mu,Y)$. Then there exist a constant $C$ and $\mu$-null set $N$ such that for $x\in S\setminus N$ the following hold $$ |f_k(x)|\leq||f_k||_{L_\infty} \leq C < \infty\;\; \text{for all k},\\ |f_k(x)-f_l(x)|\leq ||f_k-f_l||_{L^{\infty}} \rightarrow 0\;\; \text{ as}\;\; k,l \rightarrow \infty. $$ Hence for $x\in S$ there exists a function f $$ f(x):= \lim_{k\rightarrow \infty} f_k(x) \;\;\text{in} \;\;Y \;\;\text{for} \;\; x\in S\setminus N, \\ f(x):= 0 \;\;\text{for} \;\; x \in N. $$ The f is measurable and bounded, and for $x\in S\setminus N$ $$ |f(x)-f_k(x)|=\lim_{l\rightarrow \infty}|f_l(x)-f_k(x)| \leq \liminf_{l\rightarrow\infty}||f_l-f_k||_{L^{\infty}},\;\; (*) $$ and hence $$ ||f-f_k||_{L^\infty}\leq\liminf_{l\rightarrow\infty}||f_l-f_k||_{L^\infty}\rightarrow 0\;\; \text{as}\;\; k\rightarrow\infty. $$
(End of proof, end of citation)
My problem is in line labeled by (*). I agree that at the moment of writing the last inequality I am not sure about how many cluster-points $||f_l-f_k||$ (for fixed k) has. So just $\lim$ is not appropriate. I am sure it is bounded from below by 0 and from above I can bound it by arbitrary $\epsilon >0$, provided I choose both $k,l \geq n$ for $n$ large enough, because $(f_k)$ is Cauchy sequence. So both $\liminf, \limsup$ of $||f_l-f_k||$ ($k$ fixed) are finite and less than $\epsilon$. While use of limsup seems to be adequate and easy $|f_l(x)-f_k(x)|\leq\sup_{i\geq l}||f_i-f_l||_{L^{\infty}}$ and than take a limit. Similar inequality for inf simply does not hold. Of course, finally at the end, when I know there is just one cluster point, I can argue that it doesn't matter what I chose. But it can be hardly deduced at time if writing the(*). Am I correct or I am missing something? (it may be the third time I came in similar issues of using liminf while reading the book, and I am just at the beginning)
As long as $x \in S \setminus N$ you have pointed out that $|f_l(x) - f_k(x)| \le \|f_l - f_k\|_{L^\infty}$ for all $l$ and $k$.
Now fix $k$ and compute the $\liminf$ on each side. It isn't difficult to see that $\liminf$ preserves order so that $$\liminf_{l \to \infty} |f_l(x) - f_k(x)| \le \liminf_{l \to \infty}\|f_l - f_k\|_{L^\infty}.$$ But $f_l(x) \to f(x)$ implies that $|f_l(x) - f_k(x)| \to |f(x) - f_k(x)|$ so that $$\liminf_{l \to \infty} |f_l(x) - f_k(x)| = \lim_{l \to \infty} |f_l(x) - f_k(x)| = |f(x) - f_k(x)|.$$