6

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.

Victor Ronchim
  • 1,153
  • 5
  • 17
  • Because it is true for all $\mu_n$.. – Berci Sep 25 '18 at 16:54
  • @Berci, sorry. Can you be a little more specific? English is not my mother language. Is there any flaws in my argument? – Victor Ronchim Sep 25 '18 at 17:03
  • 1
    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
  • @Berci I am trying to prove the convergence $\mu_n \to \nu$. There is an easy argument for that? – Victor Ronchim Sep 25 '18 at 17:27
  • 1
    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
  • 1
    Addictive or additive? Nice typo :) – daw Sep 25 '18 at 19:13
  • @Driver8, I am not sure if it works. Could you write it? You are using $|\cdot|_\infty$ to prove the convergence. What happens when you took supremum over the measurable sets? You need to do it before taking the limit over n, right? – Victor Ronchim Sep 25 '18 at 19:14
  • @daw, they are both additive and addictive. This is a good typo indeed :) – Victor Ronchim Sep 25 '18 at 19:18
  • 1
    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

1 Answers1

1

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

Victor Ronchim
  • 1,153
  • 5
  • 17