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Let $\mathcal{M}$ be a von Neumann algebra. Then the following form of "monotone convergence theorem" holds:

If $(a_i)$ is a net of self-adjoint elements in $\mathcal{M}$ which is increasing and norm-bounded, then $a_i$ strongly converges to some $a\in\mathcal{M}$.

On the other hand, we already know that $L^\infty(0,\infty)$ is a von Neumann algebra (acting on the Hilbert space $L^2(0,\infty)$). Thus the above monotone convergence theorem would be valid in $L^\infty(0,\infty)$. In this case, we have:

If $(f_i)$ is a uniformly bounded increasing net of real measurable functions in $(0,\infty)$, then there is $f\in L^\infty(0,\infty)$ such that $$\lim_i\int_0^\infty f_ig=\int_0^\infty fg $$ holds for all $g\in L^2(0,\infty)$ with compact support.

This looks like a net version of the Lebesgue monotone convergence theorem. My question is: can we describe $f$ explicitly in terms of the $(f_i)$? [For example, "$f$ is a pointwise limit of $(f_i)$" is not the right answer, as this example shows.]

MintChocolate
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  • This is somewhat connected to this question, though the argument would need to be more elaborate in this case. The point is that the supremum of such an uncountable net can actually be reduced to a countable supremum, in which case you can just take supremum pointwise. Whether that counts as an explicit description of $f$, though, is up to interpretation. – David Gao Oct 25 '24 at 16:48

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The supremum is described in this answer. Basically, one takes the set of all finite maxima $f_{j_1}\vee\cdots\vee f_{j_r}$, and out of these one chooses the pointwise supremum of a sequence such that their integrals converge to $\sup_j\int f_j$.

The argument works in $L^\infty(X)$ for any $\sigma$-finite measure space $X$ (more generally, localizable, as MaoWao mentioned; I recommend looking at this answer). The supremum may fail to exist when $X$ is not $\sigma$-finite (an example is shown in the aforementioned answer).

Martin Argerami
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    The class of measure spaces for which the supremum of a bounded subset of $L^\infty$ always exists are called localizable measure spaces (to be precise, the definition of localizability also assumes semifiniteness of the measure). Every $\sigma$-finite measure space is localizable, but you can also encounter localizable measure spaces that are not $\sigma$-finite in everyday life, for example every uncountable set with the counting measure (on the power set) or the Haar measure on a locally compact group (with the appropriate definition). – MaoWao Oct 29 '24 at 07:23
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    I mention this because localizable measure spaces don't seem to be as well-known as they should be. They are really the right class of measure spaces for von Neumann algebras: The multiplication operator representation $L^\infty(X,\mu)\to\mathbb B(L^2(X,\mu))$ is a faithful representation onto a concrete von Neumann algebra if and only if $(X,\mu)$ is localizable. – MaoWao Oct 29 '24 at 07:26
  • Very good point. I had completely forgotten about it, and I knew it from this very site (though I cannot find the source now). – Martin Argerami Oct 29 '24 at 11:32