# The total variation norm is complete.

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.

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.