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Let $X$ be a locally compact Hausdorff space which is also $\sigma$-compact, and let $M(X)$ be the vector space of all complex Radon measures with the total variation norm $\|\mu\|:=|\mu|(X)$. Show that $M(X)$ is a Banach space.

I know that we can invoke the Riesz representation theorem to conclude that $M(X)\equiv C_0(X\to\mathbf{C})^*$, since the dual is always a Banach space, then the claim follows. I want to prove it without using the Riesz representation theorem. Let $(\mu_n)$ be a Cauchy sequence in $M(X)$, we observe that $$|\mu_m(E)-\mu_n(E)|\leq |\mu_m-\mu_n|(E)\leq \|\mu_m-\mu_n\|$$ for any measurable set $E$, thus $(\mu_n(E))$ is a Cauchy sequence, we can then define $\mu(E):=\lim_{n\to\infty}\mu_n(E)$. But I get stuck here. I don't know how to show the following two statments

  • $\mu(E)$ is a complex Radon measure.
  • $(\mu_n)$ converges to $\mu$ in $M(X)$.

For the first statement, let $E_1,E_2,\dots$ be disjoint measurable sets of $X$, we have to show $$\mu(\bigcup_{m=1}^\infty E_m)=\sum_{m=1}^\infty \mu(E_m).$$ By the definition of $\mu$, this is equivalent to $$\lim_{n\to\infty}\sum_{m=1}^\infty\mu_n(E_m)=\sum_{m=1}^\infty\lim_{n\to\infty}\mu_n(E_m).$$

Xiang Yu
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2 Answers2

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As is often the case, it is much easier to work with the following equivalent characterization of completeness of a normed vector space:

A normed vector space $(X, \|\cdot\|)$ is complete if and only if for each sequence $(x_n)_n$ in $X$ with $\sum_n \|x_n\| < \infty$, there is some $x \in X$ with $x = \sum_{n=1}^\infty x_n = \lim_{N\to\infty} \sum_{n=1}^N x_n$.

In your case, as you noted yourself, we know that $\mu(E) := \sum_{n=1}^\infty \mu_n (E)$ converges for every Borel set $E$.

Now, let $(E_n)_n$ be disjoint and let $\varepsilon >0$. There is $N$ with $\sum_{n=N}^\infty \|\mu_n\| < \varepsilon$. Hence, $$ \bigg|\sum_\ell \mu(E_\ell) - \mu(\biguplus_\ell E_\ell)\bigg| \leq \bigg|\sum_{n=1}^N \bigg(\sum_\ell \mu_n (E_\ell) - \mu_n(\biguplus_\ell E_\ell)\bigg)\bigg| + \sum_{n=N}^\infty \bigg|\sum_\ell \mu_n (E_\ell) \bigg| + \sum_{n=N}^\infty \bigg|\mu_n (\biguplus_\ell E_\ell)\bigg| < 2\varepsilon, $$ for every $\varepsilon > 0$.

Likwise, one can show regularity of $\mu$ and $\sum_{n=1}^N \mu_n \to \mu$ with convergence in $M(X)$.

PhoemueX
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  • It seems that you interchange the two sum $\sum_{\ell}\mu(E_\ell)=\sum_{\ell}\sum_{n=1}^{\infty}\mu_n(E_\ell)=\sum_{n=1}^{\infty}\sum_{\ell}\mu_n(E_\ell)$, is the series absolutely convergent? – Xiang Yu Dec 31 '15 at 13:10
  • Yes, it is. By definition of the total variation, we have $\sum_\ell |\mu_n(E_\ell)| \leq |\mu_n|(\bigcup_\ell E_\ell) \leq |\mu_n|(X)$, so that the series over $n \in \Bbb{N}$ over the previous expression is finite, since $\sum_n |\mu_n|<\infty$. – PhoemueX Dec 31 '15 at 13:14
  • Thanks. I have checked the omitted details, everything works well. – Xiang Yu Dec 31 '15 at 13:24
  • You have a small typo (likewise, not likwise). In which textbook might one find this useful characterization? I don't remember seeing it while studying Rudin's RCA. – Olorun Dec 31 '15 at 15:39
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    @Olorun You can find it in tao's blog:https://terrytao.wordpress.com/2009/01/09/245b-notes-3-lp-spaces/ – Xiang Yu Jan 01 '16 at 05:55
  • Thanks! I also found it in Folland's Real Analysis in Chapter 5 (Theorem 5.1)! – Olorun Jan 01 '16 at 06:19
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If $X$ is a locally compact Hausdorff space, by Riesz Representation Theorem, $M(X)\cong C_0(X)^*$. Note that $C_0(X)$ is a normed vector space, then its dual space $M(X)$ is a Banach space.

Here, we do not require that $X$ is $\sigma$-compact.

Mathillda
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