# 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 $$\bigvee \mathcal{F}$$?

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

• Thank you for your answer! May I ask you also if, whenever both lattice and pointwise suprema exist, the lattice supremum is greater than the pointwise one? Or does this in general not hold? Jul 24, 2020 at 13:49
• @Giuliosky Yes, that is trivial (provided the lattice order is pointwise order). By definition, the lattice supremum is an upper bound, so the same holds for the values at every point. Jul 24, 2020 at 13:52
• Sure, I was thinking to a different situation but I should have specified it. In particular, I was thinking at the lattice of all measurable functions where the lattice is induced by the a.e. order, but I guess that considering equivalence classes should work. Jul 24, 2020 at 14:00
• For equivalence classes, the pointwise (or a.e.) supremum is not even well-defined, and if you choose representatives, it can easily be strictly bigger than the lattice supremum. For example, consider the family $(1_x)$ indexed by $x\in [0,1]$. Their pointwise supremum is $1$, but the supremum in $L^1([0,1])$ is $0$. Jul 24, 2020 at 14:15

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

• As @MaoWao points out, I think it is necessary to specify in which lattice you are taking the supremum. Of course, in your case I guess that it is the lattice of measureable functions. Jul 23, 2020 at 22:58
• My I suggest to be a bit more specific in the use of the notion "lattice supremum"? The point of your Theorem 1 is not so much that you take the supremum in a lattice but, as pointed out by @LSpice, rather in which lattice you take the supremum. For instance, Theorem 1 is not true in the lattice of measurable functions, but is true in the lattice of equivalence classes that we obtain by identifying functions that are equal almost everywhere. Jul 23, 2020 at 23:17
• @JochenGlueck, thanks for the polite correction. I couldn't tell from context which lattice it was (all the more reason to specify it, I guess!), and guessed incorrectly. Jul 23, 2020 at 23:26
• @JochenGlueck I did include a link in which I carefully explained what I mean by the lattice supremum. Yes, measurable functions are identified if they coincide a.e., but that is a most common practice. Jul 24, 2020 at 0:17
• Thank you for your clear answer! Jul 24, 2020 at 13:50