Supremum of a bounded set of real-valued functions in $L^\infty(X)$

Question

Let $(X,\Sigma,\mu)$ be a measure space and consider the algebra $L^\infty(X)$. The real-valued functions carry a canonical order, where $g\leq f$ if $g(x)\leq f(x)$ a.e. The following result holds:

Let $B\subset L^\infty(X)$ be a set of real valued functions which is bounded (that is, there exists $c>0$ with $\|f\|_\infty\leq c$ for all $f\in B$). Then $B$ admits a least upper bound $g$ in $L^\infty(X)$.

As far as I can tell, the straightforward argument that one first thinks about does not work; because one would attempt to take the pointwise supremum, which of course will be bounded but there is no obvious reason why it has to be measurable. Moreover, taking the pointwise supremum raises the issue of dealing with possibly uncountably many nullsets, which kind of puts the argument away from usual measure theory.

The reason I know the result is true is the following. For any $F\subset B$ finite, we can define $g_F=\max\{f:\ f\in F\}$. Because we are dealing with finitely many functions, $g_F\in L^\infty(X)$ and $f\leq g_F$ a.e. for all $f\in F$. Now the net $\{g_F\}_{F\in B}$ is a monotone increasing bounded net of real-valued functions. Here is where I need to appeal to higher level stuff: since $L^\infty(X)$ is a von Neumann algebra, any bounded monotone increasing net of selfadjoint elements admits a least upper bound. So there exists $g\in L^\infty(X)$ that is the least upper bound of $\{g_F\}_{F\in B}$. The reason $g$ is a least upper bound for $B$ is that if $h$ is an upper bound for $B$, then $g_F\leq h$ for all $F$, and then $g\leq h$.

And now we get to the question: is it possible to prove the existence of the least upper bound for $B$, with measure theory arguments?

The proof using von Neumann algebra ideas is not crazy hard, but it uses non-trivial results. Basically one represents $L^\infty(X)$ faithfully on $B(\ell^2(X))$ acting by left multiplication, that is $\pi(f)\eta=f\,\eta$, and considers the numerical nets $\{\langle \pi(g_F)\xi,\xi\rangle\}$ for each $\xi\in \ell^2(X)$. As these are bounded monotone nets in $\mathbb R$, they converge. This way one gets numbers $\{\langle T\xi,\xi\rangle\}$ and it is standard Hilbert space theory using polarization that this defines an operator $T\in B(\ell^2(X))$. The operator $T$ is a weak-operator limit of elements of $\pi(L^\infty(X))$, which is a von Neumann algebra, and so $T\in \pi(L^\infty(X))$, which means that there exists $g\in L^\infty(X)$ with $\pi(g)=T$.

It would be nice to have instead a proof intrinsic to $L^\infty(X)$.

Answer 1

There is a counter example if you do not assume that the measure space is $\sigma$-finite. Let $X$ be an uncountable set, $\mu$ a counting measure, and $\Sigma$ the family of set $A$ such that $A$ or the complement of $A$ is countable. Consider that $S$ is a subset of $X$ such that both itself and its complement are uncountable, then $B: =\{\chi_A|A \mbox{ is a countable subset of }S\}$ has no least upper bound.

If you assume that the measure space is $\sigma$-finite, then the result you quote is true. Define $B^\vee:=\{\sup A| A\mbox{ is a finite subset of } B\}$, then $B^\vee$ is also bounded above. The key point is that the supremum $\alpha$ of integrals of elements in $B^\vee$ exists when $\mu$ is a finite measure, so we can find a sequence $(f_n)$ in $B^\vee$ such that $\int f_n$ converges to $\alpha$. Moreover, we may assume that $(f_n)$ is increasing by replacing $f_n$ with $f_1\vee f_2\vee\cdots\vee f_n$. Let $g$ be the supremum of the sequence $f_n$, then $g$ is the least upper bound of $B$. Indeed, for any $f\in B$,
\begin{align*} \alpha \geq \int (f_n\vee f)\geq \int f_n, \end{align*} which follows that $\int (f_n\vee f)$ converges to $\alpha$. By Monotone Convergence Theorem, $\int (g\vee f)-\int g=\alpha-\alpha=0$. This implies that $g\vee f-g$ is the zero element in $L^\infty(X)$, i.e., $f\leq g$.