# Countably additive finite signed measures form a Banach Space.

I'm currently studying some topics in measure theory and I am not sure how to prove the following:

Let $$X$$ a set, $$\mathcal A$$ a $$\sigma$$-algebra on X. Consider the set: $$ca(\mathcal A) = \{\mu:\mathcal A \to \mathbb R|\; \mu \; \text{is a finite signed measure} \}$$

Note that $$ca(\mathcal A)$$ is a subspace of $$l_\infty(\mathcal A)$$. I want to prove the following:

1. $$\|\mu\|\stackrel{def}{=} |\mu|(X)$$ defines a norm on $$ca(\mathcal A)$$;
2. $$(ca(\mathcal A), \| \cdot \|)$$ is a Banach Space
3. The following inequality holds: $$\|\mu\|_\infty \leq \|\mu\| \leq 2\|\mu\|_\infty,$$ where $$\|\mu\|_\infty\stackrel{def}{=} \sup\limits_{A\in \mathcal A} |\mu(A)|$$

I was able to prove (1) using Hahn-Jordan decomposition and the inequality on (3) is straightforward. Although I could not prove (2).

What I tried:

If $$(\mu_n)_{n\in \mathbb N}$$ is Cauchy sequence on $$ca(\mathcal A)$$, it follows from (3) that $$(\mu_n)$$ is point-wise convergent to $$\nu\stackrel{def}{=} \lim \mu_n$$:

Given $$\varepsilon >0$$, there is $$n_0 \geq 1$$ such that: $$m>n \geq n_0 \implies \|(\mu_m - \mu_n)\|_\infty\leq \|\mu_m - \mu_n\| <\varepsilon.$$

Hence, for all $$A\in \mathcal A$$, $$(\mu_n(A))$$ is convergent and $$\nu$$ is well defined.

I am not sure how to prove that $$\nu$$ is a signed measure in $$ca(\mathcal A)$$:

1. $$\nu(\emptyset)=\lim\mu_n(\emptyset) = 0$$;
2. Since $$(\mu_n(X))$$ is convergent sequence on $$\mathbb R$$, it is bounded. So $$\nu(X) = \lim \mu_n(X)$$ is finite.

Why, given a family $$(A_i)_{i \in \mathbb N}\subset \mathcal A$$ of disjoint sets, we have: $$\nu (\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty}\nu(A_i)$$

Given a family $$(A_i)_{i \in \mathbb N}\subset \mathcal A$$ of disjoint sets, define $$B_i = \bigcup\limits_{j=1}^i A_j$$. The family of $$(B_i)$$ is increasing, $$\cup B_i = \cup A_i$$, and:

\begin{align*} \nu(\bigcup_{i=1}^\infty A_i) &= \nu(\bigcup_{i=1}^\infty B_i) \\ &= \lim_{i} \nu(B_i)\\ &= \lim_{i} \lim_{n} \mu_n (\bigcup_{j=1}^i A_j)\\ &= \lim_{i} \lim_{n} \sum_{j=1}^i \mu_n(A_j)\\ &= \lim_{i} \sum_{j=1}^i \lim_{n} \mu_n(A_j) \\ &= \sum_{i=1}^\infty \nu(A_i). \end{align*}

Can anyone check if this is correct?

EDIT2: I forgot to check the convergence $$\mu_n \stackrel{\|\cdot\|}{\to}\nu$$ and I struggling with it.

• Because it is true for all $\mu_n$.. – Berci Sep 25 '18 at 16:54
• That's correct. However, we already know $\nu(\cup_i A_i) \ \leftarrow \mu_n(\cup_i A_i) =\sum_i\mu_n(A_i) \ \to\sum_i\nu(A_i)$.. – Berci Sep 25 '18 at 17:20
• Take $A\in \mathscr A$ and note that $|\mu_m(A)-\mu_n(A)|<\epsilon$ for sufficiently large $n,m$. Fix $n$ and let $m\to \infty.$ Then $|\nu(A)-\mu_n(A)|<\epsilon$ – Matematleta Sep 25 '18 at 18:54
• Addictive or additive? Nice typo :) – daw Sep 25 '18 at 19:13
• For every $A\in \mathscr A$ you already showed that $\mu_n(A)\to v(A).$ The function $f(x)=|x-\mu_n(A)|$ is continuous, so with $x_m=\mu_m(A)\to \nu(A)$ we have $f(x_m)\to f(\nu (A)=|\nu(A)-\mu_n(A)|.$ – Matematleta Sep 25 '18 at 19:22

The argument of $$\sigma$$-additivity for $$\nu$$ in the question is wrong. To show that $$\nu$$ is $$\sigma$$ additive, I will prove the following:

(1) If $$(A_n)_{n\in \mathbb N}\subset \mathcal A$$ is a sequence such that $$A_n\searrow \emptyset$$, then $$\nu(A_n)\to 0$$.

Given $$\varepsilon >0$$ there is a $$\mu_{m}$$ such that $$\|\nu - \mu_{m}\|_\infty<\varepsilon$$. Since $$\mu_m$$ is a measure, $$\mu_{m}(A_n) \stackrel{n}{\to} 0$$. Hence, there is $$n_0\in \mathbb N$$ such that: $$n\geq n_0 \implies |\mu_m(A_n)| < \varepsilon$$.

Thus: $$n\geq n_o \implies |\nu(A_n)| \leq |\nu(A_n)-\mu_m(A_n)| + |\mu_m(A_n)| <2\varepsilon$$

Since $$\nu$$ is finitely additive and satisfies (1), $$\nu$$ is upper-continuous: \begin{align*} A_n \nearrow A &\implies A\setminus A_n \searrow \emptyset \\ &\implies \nu(A) - \nu(A_n) \to 0\\ &\implies \nu(A_n) \to \nu(A). \end{align*}

Hence $$\nu$$ is $$\sigma$$-additive. The convergence $$\mu_n \to \nu$$ follows from the fact that $$(\mu_n)$$ is also a Cauchy sequence in the norm $$\|\cdot\|_\infty$$.