Prove that the vector measure induced by a right-continuous function is $\sigma$-additive Let


*

*$a,b\in\mathbb R$ with $a<b$ and $$\mathcal H:=\left\{(s,t]:a\le s\le t\le b\right\}$$

*$\mathcal R$ denote the ring generated by $\mathcal H$

*$E$ be a $\mathbb R$-Banach space

*$g:[a,b]\to E$ and $$\mu((s,t]):=g(t)-g(s)\;\;\;\text{for }a\le s\le t\le b$$


Note that $\mu$ is an additive set function on $\mathcal H$, i.e. $$\mu\left(\biguplus_{i=1}^nA_i\right)=\sum_{i=1}^n\mu(A_i)\tag1$$ for all disjoint $A_1,\ldots,A_n\in\mathcal H$ with $n\in\mathbb N$ and $\biguplus_{i=1}^nA_i\in\mathcal H$.

I want to show that if $g$ is right-continuous, then $\mu$ is a $\sigma$-additive set function on $\mathcal H$, i.e. $$\mu\left(\biguplus_{n\in\mathbb N}A_n\right)=\sum_{n=1}^\infty\mu(A_i)\tag2$$ for all disjoint $(A_n)_{n\in\mathbb N}\subseteq\mathcal R$ with $\biguplus_{n\in\mathbb N}A_n\in\mathcal H$.

By definition of $\mathcal H$, it's clear that the condition on $(A_n)_{n\in\mathbb N}$ implies $$A_n=(s_n,t_n]\tag 3$$ for some $a\le s_n\le t_n\le b$ for all $n\in\mathbb N$ and $$(s_m,t_m]\cap(s_n,t_n]=\emptyset\;\;\;\text{for all }m\ne n\;.\tag4$$

If we were allowed to rearrange and assume that $$s_n\le t_n\le s_{n+1}\;\;\;\text{for all }n\in\mathbb N\;,\tag5$$ then $(t_n)_{n\in\mathbb N}$ would be nondecreasing with $$t_\infty:=\lim_{n\to\infty}t_n=\sup_{n\in\mathbb N}t_n\in[s_1,b]\tag6$$ and hence (since $g$ is right-continuous)
\begin{equation}\begin{split}\mu\left(\biguplus_{n\in\mathbb N}A_n\right)&=\mu((s_1,t_\infty])=g(t_\infty)-g(s_1)\\&=\lim_{N\to\infty}g(t_N)-g(s_1)=\lim_{N\to\infty}\sum_{n=1}^N(g(t_n)-g(s_1))\\&=\sum_{n=1}^\infty\mu(A_n)\end{split}\tag7\end{equation}

This rearrangement is allowed, for example, in the classical case where $E$ is replaced by $[0,\infty]$. However, in our situation we don't know whether $\sum_{n=1}^\infty\mu(A_n)$ is absolutely convergent (or even summable).

So, how can we proved the claim here?

 A: Not that I'm going to give a counterexample, but I seriously doubt that this is true without more hypotheses; I'm going to assume that $g$ has bounded variation.
And I'm going to switch left and right, letting $H$ be the collection of half-open intervals $[a,b)$ and assuming that $g$ is left-continuous; I'll explain why later.
If $I$ and $J$ are disjoint intervals we'll say $I<J$ if $x<y$ for $x\in I$ and $y\in J$.
Suppose that $C\subset H$ is a pairwise disjoint collection of intervals with union $[0,1)$.
First, about that rearrangement that supposedly works in the classical case: The problem with the rearrangement is not just that the rearrangement might change the sum; in fact the supposed rearrangement need not exist! Because $C$ simply need not be an increasing sequence of intervals. For example, $C$ could consist of an increasing sequence of intervals covering $[0,1/2)$ together with an increasing sequence of intervals covering $[1/2,1)$.
Once you see that example you realize that the intervals in $C$ can cover $[0,1)$ in more complicated ways. In fact Folland, in a note regarding the proof he gives for the classical case in Real Analysis, comments that $(C,<)$ can be order-isomorphic to any countable ordinal. His point is that this is why the proof of the lemma is a little intricate (I think he says "fussy"); in particular it can't be done just by taking that "rearrangement". But when I saw that comment I said to myself hmm, countable ordinal...
Here's a sketch of a proof - I believe all the omitted details are straightforward. First,


$(C,<)$ is well-ordered.


Suppose not. Then there exist $I_n\in C$, $n=1,2,\dots$, with $I_{n+1}<I_n$. Now the intervals $I_n$ decrease to a point $x\in[0,1)$. We have $x\in J=[a,b)\in C$. Hence $I_n\subset J$ for large $n$, contradicting disjointness of $C$.
So $(C,<)$ is isomorphic to a countable ordinal: We have a countable ordinal $\alpha$ and $C=\{I_\beta\,:\,\beta<\alpha\}$, with $I_\beta<I_\gamma$ if and only if $\beta<\gamma$. (This is why I swapped left and right: In the original formulation $(C,<)$ would be the order-reversal of a countable ordinal.)
Another definition: If $S$ is any set and $f:S\to E$ we say $\sum_{x\in S}f(x)=s$ if for every $\epsilon>0$ there exists a finite set $F\subset S$ such that $$||s-\sum_{x\in F'}f(x)||<\epsilon$$for every finite set $F'$ with $F\subset F'\subset S$. (Note it's easy to see that if $S$ is countable and the sum converges in this sense then it also converges in thhe more usual sense, using an ordering of $S$ as a sequence.)
We need to show that $$\mu([0,1))=\sum_{\beta<\alpha}\mu(I_\beta).$$In fact it's straightforward to show that$$\mu\left(\bigcup_{\beta<\gamma}I_\beta\right)=\sum_{\beta<\gamma}\mu(I_\beta)\quad(\gamma\le\alpha)$$by transfinite induction on $\gamma$.
(The fact that $g$ is left-continuous is used for limit ordinals $\gamma$. It may well be that we don't need to assume $g$ has bounded variation to show $\mu([0,1))=\sum_{\beta<\alpha}\mu(I_\beta)$, at least if we defined that sum in the obvious way by transfinite recursion. But this is not what we actually want to prove. We also have $C=\{J_n\,:\,n\in\mathbb N\}$, and what we actually want is $\mu([0,1))=\lim_{N\to\infty}\sum_{n=1}^N\mu(J_n).$ That's clear if we define $\sum_{\beta<\alpha}$ as above instead of by recursion, but then bounded variation comes in.)
