Let $\Omega$ be a set $\mathcal A\subseteq2^\Omega$ with $\emptyset\in\mathcal A$, $E$ be a $\mathbb R$-Banach space and $\mu:\mathcal A\to E$ be additive. Now, for $A\subseteq\Omega$, let $$|\mu|(A):=\sup\left\{\sum_{i=1}^k\left\|\mu(A_i)\right\|_E\right\},$$ where the supremum is taken over all $k\in\mathbb N$ and mutually disjoint $A_1,\ldots,A_k\in\mathcal A$ with $\bigcup_{i=1}^kA_i\subseteq A$.

It's easy to see that $$\mu\mapsto|\mu|(\Omega)\tag1$$ is a norm on the vector space of those $\mu$ for which $|\mu|(\Omega)<\infty$. Are we able to show that this norm is complete?

Assuming $(\mu_n)_{n\in\mathbb N}$ is a Cauchy sequence wrt $(1)$ of such $\mu$. For $\varepsilon>0$, there is a $N\in\mathbb N$ with $$|\mu_m-\mu_n|(\Omega)<\varepsilon\;\;\;\text{for all }m,n\ge N\tag2.$$ We should clearly have $$\left||\mu_m|(A)-|\mu_n|(A)\right|\le|\mu_m-\mu_n|(A)\le|\mu_m-\mu_n|(\Omega)\tag3.$$ So, $$(|\mu_n|(A))_{n\in\mathbb N}$$ is Cauchy.

If $E=\mathbb R$ this might help (a signed measure can be decomposed into a negative and a positive part), but I don't see what we need to do in general.

BTW: It would be great if someone knows a textbook reference where it is shown that the space of $E$-valued vector measures equipped with the total variation norm is complete.

  • $\begingroup$ Do the inequalities in (3) require a decomposition theorem? It looks like they should be straightforward. $\endgroup$
    – Umberto P.
    Apr 22, 2019 at 17:24

1 Answer 1


This isn't too different from the proofs that any of the basic function spaces are complete. Indeed suppose that $(\mu_n)_{n \geq 1}$ is Cauchy for $| \cdot |(\Omega)$. Let $A_i \in \mathcal{A}$. Then $$\| \mu_n(A_i) - \mu_m(A_i) \| \leq | \mu_n - \mu_m |(\Omega)$$ so $\mu_n(A_i)$ is a Cauchy sequence in the Banach space $E$ and hence converges to some element of $E$, which we will call $\mu(A_i)$. So we now have a function $\mu: \mathcal{A} \to E$ which is the pointwise limit of the sequence $\mu_n$. It is hence immediate that $\mu$ is finitely additive.

It remains to see that $|\mu|(\Omega) < \infty$ and that $|\mu_n - \mu|(\Omega) \to 0$. Let's prove them in that order. For the first, take arbitrary disjoint $A_1, \dots A_k \in \mathcal{A}$. Since $\mu_n \to \mu$ pointwise, for $n$ large enough $\sum_i \| \mu_n(A_i) - \mu(A_i) \| \leq 1$. Then, we can estimate, $$\sum_{i=1}^k \|\mu(A_i)\| \leq 1 + \sum_i \|\mu_n(A_i)\| \leq 1 + \sup_n |\mu_n|(\Omega) < \infty$$ since Cauchy sequences are bounded. So taking the $\sup$ on the left hand side gives us that $| \mu |(\Omega) \leq 1 + \sup_n |\mu_n|(\Omega) < \infty$.

Finally, to see that $\mu_n \to \mu$ for your norm, note that for $\varepsilon > 0$ there is an $N$ (independent of our choice of $A_i$) such that for $n,m \geq N$, $$\sum_i \|\mu_n(A_i) - \mu_m(A_i)\| \leq |\mu_n - \mu_m|(\Omega) \leq \varepsilon$$ Sending $m \to \infty$ on the left hand side and taking the $\sup$ gives for $n \geq N$ $$|\mu_n - \mu|(\Omega) \leq \varepsilon$$ which shows the desired convergence.

  • $\begingroup$ In $\sum_{i=1}^k \|\mu(A_i)\| \leq 1 + \sum_i \|\mu_n(A_i)\| \leq 1 + \sup_n |\mu_n|(\Omega) < \infty$, how did you obtain the last inequality? Clearly, there is a $N\in\mathbb N$ with $\sum_{i=1}^k|\mu_N(A_i)-\mu(A_i)|<1$. Then $\sum_{i=1}^k|\mu(A_i)|\le\sum_{i=1}^k|\mu_n(A_i)-\mu(A_i)|+\sum_{i=1}^k|\mu_N(A_i)|<1+k|\mu_N(\Omega)|$. But this tends to $\infty$ as $k\to\infty$. $\endgroup$
    – 0xbadf00d
    Jul 18, 2019 at 6:29
  • $\begingroup$ $\sum_i |\mu_N(A_i)| \leq |\mu_N|(\Omega)$ without the $k$ there. This is literally just the definition of $|\mu_N|$. $\endgroup$ Jul 18, 2019 at 9:42
  • $\begingroup$ Sure I forgot the definition for a moment. If each $\mu_n$ is even countable additive, is $\mu$ countable as well? $\endgroup$
    – 0xbadf00d
    Jul 18, 2019 at 11:06
  • $\begingroup$ $\mu$ is certainly countably additive if you're willing to assume that $\mathcal{A}$ is a $\sigma$-algebra. The argument I know does use that $\mathcal{A}$ is closed under complements so if you're not willing to assume that I'd have to think. Either way, the argument is too long to reproduce in the comments here. $\endgroup$ Jul 18, 2019 at 11:50
  • 1
    $\begingroup$ That's basically the same as the proof I had in this case, so yep seems good to me :) $\endgroup$ Jul 18, 2019 at 13:10

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