Setwise limit of measures is a measure, is that true for nets as well?

Question

It is well known that the setwise limit of a sequence of measures is a measure.

Is the same true for nets? (Note that the proof given in the link above relies crucially on Radon-Nikodym, and so it only works for sequences.)

A reference would also be welcome.

Answer 1

The result of the MSE post you linked can indeed be generalized to nets.

Let $(\nu_\iota)_{\iota\in I}$ be a setwise convergent net of complex measures on a $\sigma$-algebra $\mathcal B$. Suppose that there exists a finite positive measure $\mu$ on $\mathcal B$ as well as an index $\kappa\in I$ such that $|\nu_\iota(B)|\leq\mu(B)$ for every $\iota\in I$ such that $\iota\geq\kappa$ and every $B\in\mathcal B$. Then, the setwise limit $\nu$ of the net $(\nu_\iota)_{\iota\in I}$ is a complex measure on $\mathcal B$.

The set function $\nu$ is obviously finitely additive and satisfies $|\nu(B)|\leq\mu(B)$ for every $B\in\mathcal B$. It follows that $\lim_{n\to\infty}\nu(B_n)=0$ for every decreasing sequence $(B_n)_{n\in\mathbb N}$ of elements of $\mathcal B$ with empty intersection. From this property combined with finite additivity, it can be checked that $\nu$ is countably additive. Indeed, if $(A_n)_{n\in\mathbb N}$ is a sequence of pairwise disjoint sets in $\mathcal B$ with union $A$, then $$\sum_{n\in\mathbb N}\nu(A_n)=\lim_{n\to\infty}\nu\mathopen{}\left(\bigcup_{k=0}^nA_k\right)=\lim_{n\to\infty}\nu\mathopen{}\left(A\setminus\bigcup_{k>n}A_k\right)=\lim_{n\to\infty}\left(\nu(A)-\nu\mathopen{}\left(\bigcup_{k>n}A_k\right)\right)=\nu(A)$$ because $(\bigcup_{k>n}A_k)_{n\in\mathbb N}$ is a decreasing sequence of elements of $\mathcal B$ with empty intersection.

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Remark: If $\mu$ is a finite positive measure on $\mathcal B$ and $\lambda$ is a complex finitely additive function on $\mathcal B$ that is absolutely continuous with respect to $\mu$ (in the sense of "for every $\varepsilon>0$, there is a $\delta>0$ such that for every $B\in\mathcal B$, the relation $\mu(B)\leq\delta$ implies $|\lambda(B)|\leq\varepsilon$"), it is once again true that $\lim_{n\to\infty}\lambda(B_n)=0$ for every decreasing sequence $(B_n)_{n\in\mathbb N}$ of elements of $\mathcal B$ with empty intersection.

So you can get more general results such as

Let $(\nu_\iota)_{\iota\in I}$ be a setwise convergent net of complex measures on a $\sigma$-algebra $\mathcal B$. Suppose that there exists a finite positive measure $\mu$ on $\mathcal B$ satisfying the following condition: $$\forall\varepsilon>0,~\exists\kappa\in I,~\exists\delta>0,~\forall\iota\in I,~\forall B\in\mathcal B,\qquad(\iota\geq\kappa\quad\text{and}\quad\mu(B)\leq\delta)\implies|\nu_\iota(B)|\leq\varepsilon.$$ Then, the setwise limit $\nu$ of the net $(\nu_\iota)_{\iota\in I}$ is a complex measure on $\mathcal B$.

The condition written in quantifiers here is pretty much what is obtained during the proof of the Vitali-Hahn-Saks theorem presented on Wikipedia. A practical situation in which this condition is satisfied is when there is an index $\kappa\in I$ such that $\nu_\iota\ll\mu$ for every $\iota\geq\kappa$ and such that the family $(d\nu_\iota/d\mu)_{\iota\geq\kappa}$ is uniformly integrable with respect to $\mu$.

Answer 2

Another way to see that the proof you linked does generalize is that in general, if a net $\nu_\alpha$ converges to some $\nu$ in the topology of setwise convergence (i.e., the product topology on $\mathbb R^{\mathcal B}$), then for every countable family $A_i$ there is a sequence $\nu_{\alpha_k}$ which converges to $\nu$ on each $A_i$.

To see this, just observe by definition of convergence in a net there is an index $\alpha_1$ of the net after which we always have $|\nu_\alpha(A_1)-\nu(A_1)|<1$, then a later index $\alpha_2$ after which we have $|\nu_\alpha(A_1)-\nu(A_1)|<\frac{1}{2}$, $|\nu_\alpha(A_2)-\nu(A_2)|<\frac{1}{2}$, and so on for every $k$, an index $\alpha_k\geq \alpha_{k-1}$ after which $|\nu_\alpha(A_i)-\nu(A_i)|<\frac{1}{n}$ for each $i\leq n$.

Therefore we as good as do have a setwise convergent sequence, as it converges on the sets of interest in the proof.