lattice suprema vs pointwise suprema What is the difference between the lattice supremum and the pointwise supremum of a family of functions? I mean, given a family of real valued functions $\mathcal{F}$, is the function $\sup\mathcal{F}:x\mapsto \sup\left\lbrace f(x):f\in \mathcal{F}\right\rbrace$ different from $\bigvee \mathcal{F}$?
 A: The answer also depends on the Banach lattice (even if the lattice order is pointwise order). If the functions from $\mathcal{F}$ are pointwise bounded, then the pointwise supremum exists, but the supremum in the lattice may not.
For example, if the lattice is $C([0,1])$ and $f_n(x)=\max\{1-(2x)^n,0\}$, then the pointwise supremum is $f(x)=1_{[0,1/2)}$, but the family has no supremum in $C([0,1])$.
Even if the supremum in the lattice exists, it does not have to coincide with the pointwise supremum. As an example, you can again take $C([0,1])$ as Banach lattice and $f_n(x)=1-x^n$. Then the pointwise supremum is $1_{[0,1)}$, but the supremum in $C([0,1])$ is $1_{[0,1]}$.
A: 
Theorem. Let $\mathcal{F}$ be a class of measurable functions defined in a
measurable set $E\subset\mathbb{R}^n$. Then $\bigvee\mathcal{F}$
exists and there is a countable subfamily
$\mathcal{G}\subset\mathcal{F}$ such that $$
 \bigvee\mathcal{F}=\bigvee \mathcal{G}=\sup \mathcal{G}. $$
In particular the lattice supremum of an uncountable family of measurable functions is measurable.

However, a pointwise supremum need not be measurable as the following example shows:

Example. Let $I\subset[0,1]$ be a non-measureable set and for $i\in I$ we define $$ f_i(x)= \begin{cases} 1 & \text{if $x=i$},\\ 0 &
 \text{if $x\neq i$.} \end{cases} $$ Then all functions $f_i$ are Borel
measurable, however, $\sup_{i\in I} f_i=\chi_I$ is the characteristic
function of a non-measurable set and hence is non-measurable.

For a detailed definition of the lattice supremum and a detailed proof of the above theorem, see
https://mathoverflow.net/a/316658/121665
