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I'm trying to proof that the total variation norm is complete. I defined the total variation associated to a signed measure $\mu$ (in the signed measure space $(X,\Sigma,\mu)$) as the map $|\mu|:\Sigma\rightarrow\overline{\mathbb{R}}$ such that:

$$|\mu|(A)=\sup\left\{\sum_{n=1}^{\infty}{|\mu(A_n)|}\middle|\{A_n\}_{n\in\mathbb{N}}\subset\Sigma\text{ is a partition of }A\right\}.$$

It is proven that $|\mu|$ is a measure. Furthermore, $|\mu|$ iff $\text{im}(\mu)\subset\mathbb{R}$. We say that a signed measure space $(X,\Sigma,\mu)$ is finite if $|\mu|$ is finite. In this case, we define the total variation norm of $\mu$ as the following value:

$$||\mu||=|\mu|(X).$$

It is easy to demonstrate that $||·||$ is a norm on the vector space $M(X,\Sigma)$ of finite signed measures. My only goal is to prove that is complete.

First, I chose a Cauchy's sequence $\{\mu_n\}_{n\in\mathbb{N}}\subset M(X,\Sigma)$. The fact that $|\mu_m(\cdot)-\mu_n(\cdot)|\leq|\mu_m-\mu_n|(\cdot)$ in $\Sigma$, implies that the sequence $\{\mu_n(A)\}_{n\in\mathbb{N}}\subset\mathbb{R}$ is Cauchy's forall $A\in\Sigma$, thus convergent. We can define $\mu$ as the pointwise limit of $\mu_n$. That convergence is actually uniform because of the inequality $|\mu_m(\cdot)-\mu_n(\cdot)|\leq|\mu_m-\mu_n|(\cdot)$. I've proved that $\mu$ is actually a finite signed measure.

My point is, how can I prove that $||\mu_n-\mu||\stackrel{n\to\infty}{\rightarrow}0$. Because of the inequality that I mentioned before, fixed $\varepsilon>0$, there exists $k_0=k_0(\varepsilon)\in\mathbb{N}$ such that for all $k,l\geq k_0$, we have:

$$\sum_{n=1}^{\infty}{|\mu_k(A_n)-\mu_l(A_n)|}\leq||\mu_k-\mu_l||<\dfrac{\varepsilon}{2},\forall\{A_n\}_{n\in\mathbb{N}}\subset\Sigma\text{ partition of }X.$$

If I can interchange the limit as $l\to\infty$ with the series, the prove is done, because all I need next is to take the supremum over all partitions of $X$. But how can I be sure that I can do that? I want to prove that I'm able to do that.

Thanks in advance.

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It is very simple ! Let $N$ be any positive integer.

$$\sum_{n=1}^{N}{|\mu_k(A_n)-\mu_l(A_n)|}\leq||\mu_k-\mu_l||<\dfrac{\varepsilon}{2},\forall\{A_n\}_{n\in\mathbb{N}}\subset\Sigma\text{ partition of }X.$$ Now let $l \to \infty$ to get $$\sum_{n=1}^{N}{|\mu_k(A_n)-\mu(A_n)|}<\dfrac{\varepsilon}{2},\forall\{A_n\}_{n\in\mathbb{N}}\subset\Sigma\text{ partition of }X.$$ Now you are ready to let $N \to \infty$ and finish the proof.

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  • $\begingroup$ Oh! Thank you very much! I didn't realize, hahaha. $\endgroup$ Feb 25, 2022 at 9:44

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