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Let $M(X)$ be the space of finite signed borel measures on $X$ with the norm $\|\mu\|=|\mu|(X)$, the variation of the measure. I am trying to prove that this is a Banach space by using the dominated convergence theorem. Let $\mu_n$ be Cauchy. Using the fact that $|\mu(E)|\leq|\mu|(E)$ we can define $\mu_n(E)\rightarrow\mu(E)$. Now, I am stuck at showing the countable additivity of $\mu$.

Let ${E_i}$ be disjoint, let $E=\cup E_i$. By definition $\mu(E)=lim(\Sigma\mu_n(E_i))$ To show that this is equal to $\Sigma \mu(E_i)$, I take the counting measure on the natural numbers. To use the dominated convergence theorem, I need to find a real absolutely convergent sequence ${a_i}$ such that $\mu_n(E_i)\leq a_i \forall n$. I am stuck at producing such a sequence, could someone guide me on how to continue?

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You can't in general find such a dominating sequence. Consider as an example $\mu_n = \frac{1}{n}\delta_{x_n}$, where $(x_n)$ is a sequence of distinct points in $X$. Then if $E_i = \{x_i\}$ we have

$$\sup_n \mu_n(E_i) = \frac{1}{i}$$

for every $i$, so no dominating sequence is summable. But $(\mu_n)$ clearly converges to $0$.

You need a different approach. It is easily seen that $\mu$ is finitely additive, and bounded.

Now you can use (maybe you need to first prove it) the fact that a bounded finitely additive set function $\gamma \colon\mathcal{B}(X) \to \mathbb{C}$ is $\sigma$-additive if and only if

$$\lim_{n\to \infty} \gamma(A_n) = 0$$

for all sequences $A_n \supseteq A_{n+1}$ of Borel sets with

$$\bigcap_{n = 1}^{\infty} A_n = \varnothing.$$

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  • $\begingroup$ Wow.. That was a slick counterexample.. Thanks! $\endgroup$ – Manan Jun 17 '17 at 14:56

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