The supremum of an arbitrary set of measurable functions from a $\sigma$-finite measure space into $\mathbb R\cup \{\pm\infty\}$ exists in the following sense:

Let $F$ be a set of such measurable functions. Then there is measurable $g$ such that $f\le g$ a.e. for all $f\in F$. And if $h$ is such that $f\le h$ a.e. for all $f\in F$, then $g\le h$.

The trick is that the inequalities are required in the a.e. sense. I have seen proofs that use Zorn's lemma (which is tempting), but there is a proof without it (see, e.g., Bogachev's monograph on measure theory, it uses monotone convergence). The result is also surprising because many properties in measure/integration theory have countability built-in.