• $\Omega$ be a set
  • $\mathcal A$ be a $\sigma$-algebra on $\Omega$
  • $\mu:\mathcal A\to\mathbb R$ be $\sigma$-additive

If $\mathcal S\subseteq2^\Omega$ with $\emptyset\in\mathcal S$, then $$\operatorname{Var}_\mu^{\mathcal S}(A):=\sup\left\{\sum_{i=1}^n|\mu(S_i)|:n\in\mathbb N\text{ and }S_1,\ldots,S_n\in\mathcal S\text{ are pairwise disjoint with }\biguplus_{i=1}^nS_i\subseteq A\right\}$$ for $A\subseteq\Omega$.

In Corollary 1.10 of the book Vector Measures by Diestel and Uhl, the authors seem to use the identity $$\operatorname{Var}_\mu^{\mathcal S}(S)=\operatorname{Var}_\mu^{\mathcal A}(S)\;\;\;\text{for all }S\in\mathcal S\;.\tag1$$ In their scenario, $\mathcal S$ is an algebra on $\Omega$ with $\sigma(\mathcal S)=\mathcal A$. How can we prove $(1)$?

I don't know how we need to argue, but if $(\mu^+,\mu^-)$ denotes the Jordan decomposition of $\mu$, then we clearly have $$\operatorname{Var}_\mu^{\mathcal A}(A)=(\mu^++\mu^-)(A)\;\;\;\text{for all }A\in\mathcal A\tag2\;.$$

In the same way we obtain the following: If $\mathcal S$ is $\cup$-stable, then $$\operatorname{Var}_\mu^{\mathcal S}(S)\le(\mu^++\mu^-)(S)\;\;\;\text{for all }S\in\mathcal S\tag3\;.$$ The problematic thing is the other inequality.

The other inequality in $(2)$ is proved in the following way: Let $(\Omega^+,\Omega^-)$ be a Hahn decomposition of $\Omega$ with respect to $\mu$. Then, $$A\cap\Omega^\pm\in\mathcal A$$ are disjoint with $$A=A\cap\Omega^+\uplus A\cap\Omega^-\tag4$$ and hence \begin{equation}\begin{split}(\mu^++\mu^-)(A)&=\mu(A\cap\Omega^+)-\mu(A\cap\Omega^-)\\&=|\mu(A\cap\Omega^+)|+|\mu(A\cap\Omega^-)|\\&\le\operatorname{Var}_\mu^{\mathcal A}(A)\end{split}\tag5\end{equation} for all $A\in\mathcal A$. The key point is that $(4)$ is a partition of $A$ with elements from $\mathcal A$. Since we don't know whether $A\cap\Omega^\pm\in\mathcal S$, we cannot proceed in the same way to establish $(1)$.

  • $\begingroup$ Please edit to clarify the question, it is not clear what you need to understand. $\endgroup$ – Evgeny Kuznetsov Feb 14 '18 at 20:25
  • $\begingroup$ @EvgenyKuznetsov I don't understand how we can prove "$\ge$" in $(1)$. (And as I wrote in the question, this seems to be used in the book: see here after "but for the same" ...) $\endgroup$ – 0xbadf00d Feb 14 '18 at 20:26
  • $\begingroup$ Can't you approximate partitions into pieces from $\mathcal A$ by partitions into pieces from $\mathcal A$? Then the sup in the definition of the variation would be the same. $\endgroup$ – Andreas Blass Feb 14 '18 at 22:13
  • $\begingroup$ @AndreasBlass Please be more explicit. I'm not sure what you're suggesting. $\endgroup$ – 0xbadf00d Feb 14 '18 at 23:28
  • $\begingroup$ The second $\mathcal A$ in my previous comment should have been $\mathcal S$. $\endgroup$ – Andreas Blass Feb 16 '18 at 13:31

We can prove the claim even in a more general setting. For the sake of clarity, I redeclare the objects: Let

  • $\Omega$ be a set
  • $\mathcal A$ be an algebra on $\Omega$
  • $E$ be a $\mathbb R$-Banach space
  • $\mu:\mathcal A\to E$ be bounded and $\sigma$-additive

We will need the following Lemma:

Let $\nu:\mathcal A\to[0,\infty)$ be $\sigma$-additive with $\mu\ll\nu$ and $\overline\nu$ denote the unique extension of $\nu$ to a $\sigma$-additive function $\sigma(\mathcal S)\to[0,\infty)$ $\Rightarrow$ $\mu$ has a unique extension $\overline\mu:\sigma(\mathcal A)\to E$ with $\overline\mu\ll\overline\nu$.

The proof is surprisingly simple:

  • Note that $$d_{\overline\nu}(A,B):=\overline\nu(A\:\Delta\:B)\;\;\;\text{for }A,B\in\sigma(\mathcal A)$$ is a semimetric on $\sigma(\mathcal A)$ and
  • $\mu\ll\nu$ is equivalent to $\mu$ being uniformly continuous wrt $d_\nu$
  • $\mathcal A$ is dense in $\sigma(\mathcal A)$ wrt $d_{\overline\nu}$
  • Thus, as $E$ is complete, there is a unique extension $\overline\mu:\sigma(\mathcal A)\to E$ of $\mu$ and $\overline\mu$ is uniformly continuous wrt $d_{\overline\nu}$, i.e. $\overline\mu\ll\overline\nu$
  • We can show that $\overline\mu$ is $\sigma$-additive

Now, assume $\mu$ has bounded variation, i.e. $\operatorname{Var}_\mu^{\mathcal A}(\Omega)<\infty$, and choose $$\nu:=\left.\operatorname{Var}_\mu^{\mathcal A}\right|_{\mathcal A}.$$ $\nu$ is $\sigma$-additive and $$\left\|\mu(A)\right\|_E\le\nu(A)\;\;\;\text{for all }A\in\mathcal A$$ and hence $$\mu\ll\nu.$$ By the lemma, $\mu$ has a unique $\sigma$-additive extension $\overline\mu:\sigma(\mathcal A)\to E$ with $$\overline\mu\ll\overline\nu.$$

Now, all we need to do is showing $$\left\|\overline\mu(A)\right\|_E\le\overline\nu(A)\;\;\;\text{for all }A\in\sigma(\mathcal A),$$ since then $$|\mu|(A)\le\overline\nu(A)\;\;\;\text{for all }A\in\sigma(\mathcal A)$$ and the other inequality is trivial. So, let $A\in\sigma(\mathcal A)$. Since $\mathcal A$ is dense in $\sigma(\mathcal A)$ wrt $d_{\overline\nu}$, there is a sequence $(A_n)_{n\in\mathbb N}\subseteq\mathcal A$ with $$d_{\overline\nu}(A,A_n)\xrightarrow{n\to\infty}0.$$

Since $\overline\mu\ll\overline\nu$, $\overline\mu$ is (uniformly) continuous with respect to $d_{\overline\nu}$ and hence $$\mu(A_n)=\overline\mu(A_n)\xrightarrow{n\to\infty}\overline\mu(A).$$ Clearly, $\overline\nu$ is continuous wrt $d_{\overline\nu}$ and hence $$\nu(A_n)=\overline\nu(A_n)\xrightarrow{n\to\infty}\overline\nu(A)$$ and as $$\left\|\mu(A_n)\right\|_E\le\nu(A_n)\;\;\;\text{for all }n\in\mathbb N,$$ we conclude $$\left\|\overline\mu(A)\right\|_E\le\overline\nu(A).$$

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Yes "For $\mu$ an algebra is "dense" for its $\sigma-$algebra"

We use the following proposition: For any $A\in \mathcal{A}$ and $\epsilon>0 $ there exists $S\in \mathcal{S}$ such that $$|\mu(A \Delta S)|\leq \epsilon .$$ Proof of the proposition: for $B\in \mathcal{A}$ we note $rg(B)$ the minimum number of sigma unions used to construct $B$ from $S$. Formally it is defined by $$ rg(S)=0$$ and for $B\in \mathcal{A}$ $$B=\cup_{\in \mathbb{N}}B_i $$ then $rg(B)=\max_i rg(B_i)+1$.

To prove the proposition, let $B$. let $N$ such that $\mu(B-\cup_{i\leq N}B_i)\leq \epsilon/2$ by iterative hypothesis take $S_i$ such that $\mu(B_i-S_i)\leq \frac{\epsilon}{2^{i+1}}$. Then $\mu(B-\cup_{i\leq N}S_i)\leq \epsilon$. (end of the proof.)

To conclude for $Var^\mathcal{A}$ and $Var^S$, Let $\epsilon >0$. There exist $A_1,\cdots ,A_N$ such that $Var^\mathcal{A}(S)-\sum_{i=1}^N |\mu(A_i)|\leq \epsilon$. Now, let $S_i$ such that $|\mu(A_i \Delta S_i)|\leq \epsilon/N^2 $. Call $\tilde{S}_i=S_i-S_i\cap\cup_{j\neq i}S_j$ we get $$\sum_i |\mu(\tilde{S}_i\Delta A_i)|\leq \epsilon $$ and then $$ Var^{S}(S)\geq \sum |\mu(\tilde{S}_i)|\geq \sum_i |\mu(A_i)|-\epsilon \geq Var^{\mathcal{A}}(S)-2\epsilon$$

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  • $\begingroup$ Sorry, but I don't understand your proof. It's clear to me that for any $\uplus$-stable semiring $\mathcal H\subseteq2^\Omega$, we've got $$\inf_{H\in\mathcal H}\lambda(A\:\Delta\:H)=0\;\;\;\text{for all }A\in\sigma(\mathcal H),$$ $\lambda:\sigma(\mathcal H)\to[0,\infty]$ being $\sigma$-additive and $\sigma$-finite on $\mathcal H$. But I don't see how this enables us to conclude the claim. $\endgroup$ – 0xbadf00d Feb 17 '18 at 23:00

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