lattice suprema vs pointwise suprema

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 $1_{[0,1]}$.

Theorem. Let $mathcal{F}$ be a class of measurable functions defined in a measurable set $Esubsetmathbb{R}^n$. Then $bigveemathcal{F}$ exists and there is a countable subfamily $mathcal{G}subsetmathcal{F}$ such that $$ bigveemathcal{F}=bigvee mathcal{G}=sup mathcal{G}. $$ 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 $Isubset[0,1]$ be a non-measureable set and for $iin I$ we define $$ f_i(x)= begin{cases} 1 & text{if $x=i$},\ 0 & text{if $xneq i$.} end{cases} $$ Then all functions $f_i$ are Borel measurable, however, $sup_{iin I} f_i=chi_I$ 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