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.

  • $\begingroup$ Because it is true for all $\mu_n$.. $\endgroup$ – Berci Sep 25 '18 at 16:54
  • 1
    $\begingroup$ 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) $.. $\endgroup$ – Berci Sep 25 '18 at 17:20
  • 1
    $\begingroup$ 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 $|\nu(A)-\mu_n(A)|<\epsilon$ $\endgroup$ – Matematleta Sep 25 '18 at 18:54
  • 1
    $\begingroup$ Addictive or additive? Nice typo :) $\endgroup$ – daw Sep 25 '18 at 19:13
  • 1
    $\begingroup$ 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 $f(x_m)\to f(\nu (A)=|\nu(A)-\mu_n(A)|.$ $\endgroup$ – 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 $\|\cdot\|_\infty$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.