# Space of vector measures equipped with the total variation norm is complete

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.

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

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.
• 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 $k\to\infty$. Jul 18, 2019 at 6:29
• $\sum_i |\mu_N(A_i)| \leq |\mu_N|(\Omega)$ without the $k$ there. This is literally just the definition of $|\mu_N|$. Jul 18, 2019 at 9:42
• Sure I forgot the definition for a moment. If each $\mu_n$ is even countable additive, is $\mu$ countable as well? Jul 18, 2019 at 11:06
• $\mu$ 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. Jul 18, 2019 at 11:50