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Let

  • $(\Omega,\mathcal A)$ be a measurable space
  • $E$ be a $\mathbb R$-Banach space
  • $\mu:\mathcal A\to E$ with $\mu(\emptyset)=0$ and $$\mu\left(\biguplus_{n\in\mathbb N}A_n\right)=\sum_{n\in\mathbb N}\mu(A_n)\tag1$$ for all disjoint $(A_n)_{n\in\mathbb N}\subseteq\mathcal A$

Now, let $$|\mu|(A):=\sup\left\{\sum_{i=1}^n\left\|\mu(A_i)\right\|_E:n\in\mathbb N\text{ and }A_1,\ldots,A_n\in\mathcal A\text{ are disjoint with }\biguplus_{i=1}^nA_i\subseteq A\right\}$$ for $A\in\mathcal A$. I've read in a lecture note that $|\mu|$ is a measure on $(\Omega,\mathcal A)$, if $|\mu|(\Omega)<\infty$. What goes wrong if $|\mu|(\Omega)=\infty$?

Let $(A_n)_{n\in\mathbb N}\subseteq\mathcal A$ be disjoint and $A:=\biguplus_{n\in\mathbb N}A_n$. First of all, it's easy to see (and that's even stated in that lecture note) that $$\sum_{n\in\mathbb N}|\mu|(A_n)\le|\mu|(A)\tag2\;.$$ So, the problem must occur in the proof of the other inequality:

  • Let $k\in\mathbb N$ and $B_1,\ldots,B_k\in\mathcal A$ be disjoint with $$\biguplus_{i=1}^kB_i\subseteq A\tag3$$
  • Then, $(A_n\cap B_i)_{n\in\mathbb N}$ is disjoint with $$\biguplus_{n\in\mathbb N}(A_n\cap B_i)=B_i\tag4$$ for all $i\in\left\{1,\ldots,k\right\}$ and $A_n\cap B_1,\ldots,A_n\cap B_k$ are disjoint with $$\biguplus_{i=1}^k(A_n\cap B_i)\subseteq A_n\tag5$$ for all $n\in\mathbb N$
  • Thus, \begin{equation}\begin{split}\sum_{i=1}^k\left\|\mu(B_i)\right\|_E&=\sum_{i=1}^k\left\|\mu\left(\biguplus_{n\in\mathbb N}(A_n\cap B_i)\right)\right\|_E=\sum_{i=1}^k\left\|\sum_{n\in\mathbb N}\mu(A_n\cap B_i)\right\|_E\\&\le\sum_{i=1}^k\sum_{n\in\mathbb N}\left\|\mu(A_n\cap B_i)\right\|_E\\&=\sum_{n\in\mathbb N}\sum_{i=1}^k\left\|\mu(A_n\cap B_i)\right\|_E\le\sum_{n\in\mathbb N}|\mu|(A_n)\end{split}\tag6\end{equation}
  • $(6)$ should immediately yield $$|\mu|(A)\le\sum_{n\in\mathbb N}|\mu|(A_n)\tag7$$

By $(2)$ and $(7)$ we obtain the $\sigma$-additivity of $|\mu|$. Hence, $|\mu|$ is a measure (clearly, not a finite one, but that wasn't claimed). So, is there anything wrong in my proof?

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  • $\begingroup$ Maybe this is silly, but I'm having trouble seeing how a vector measure of unbounded variation can exist at all. Is there a standard example? $\endgroup$ – Nate Eldredge Jan 20 '18 at 15:03
  • $\begingroup$ What does (1) mean exactly? A priori the sum is defined only if $\sum ||\mu(A_i)||<\infty$, right? So the $\sigma$-additivity of $\mu$ is not general. What then can follow for $|\mu|$ ? $\endgroup$ – Jean Duchon Jan 20 '18 at 15:10
  • $\begingroup$ @JeanDuchon: Yes, that's what was worrying me. $\endgroup$ – Nate Eldredge Jan 20 '18 at 15:18
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    $\begingroup$ @NateEldredge I'm sorry, you're right. Here is a working counterexample: Let $(\Omega,\mathcal A)=(\mathbb N,2^{\mathbb N})$, $(\alpha_n)_{n\in\mathbb N}\in c_0$ and $$\mu(I):=\sum_{i\in I}\alpha_ie^i\;\;\;\text{for }I\subseteq\mathbb N\;,$$ where $e^i\in C_0$ is such that $e^i_j=\delta_{ij}$. Then, $$|\mu|(I)=\sum_{i\in I}|\alpha_i|\;\;\;\text{for all }I\subseteq\mathbb N$$ and hence $\mu$ has bounded variation iff $\alpha\in\ell_1$. $\endgroup$ – 0xbadf00d Jan 20 '18 at 17:31
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    $\begingroup$ @NateEldredge Also, in any Hilbert space with an orthogonal basis $\{e_\lambda\}_{\lambda\in\Lambda}$, an element $u\in H$ is the sum of the summable family $\{(u,e_\lambda)e_\lambda\}_{\lambda\in\Lambda}$, (which of course in general is not absolutely summable). $\endgroup$ – Pietro Majer Jan 20 '18 at 17:48
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In fact the total variation of a Banach valued vector measure is always a measure, as you are saying. What can happen is that $\|\mu\|$ has "heavy atoms", that is, measurable, non null sets $A\in\mathcal{A}$ such that any measurable $B\subset A$ is either $\|\mu\|$-null or $\|\mu\|$-infinite. The space $\Omega$ itself may result a "heavy atom", so that the image of $\|\mu\|$ as a function is reduced to $\{0,+\infty\}$. Maybe the lecture notes wanted to rule out these somehow pathological situations.

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