Countably additive finite signed measures form a Banach Space.

functional-analysismeasure-theorysigned-measures

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:
$$\nu (\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty}\nu(A_i)$$

EDIT: I had an idea:

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.

Best Answer

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 $\|\cdot\|_\infty$.

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