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Let $X$ be a topological space and let $\mu$ be a Borel, regular measure, with finite total variation $\| \mu \| _{TV}$. One may view $\mu$ as a bounded linear functional on the Banach space $C_b (X)$ of the bounded continuous functions on $X$; as such, it has a norm given by $\| \mu \| = \sup _{\| f \| = 1} \left| \int _X f \ \mathrm d \mu \right|$.

Are the two norms of $\mu$ equal?

(In particular, an affirmative answer would clarify why the total variation norm has this slightly unintuitive definition.)

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  • $\begingroup$ Is $\mu$ here a signed or complex measure? It seems to be trivial if $\mu$ is a positive measure. $\endgroup$ Jan 24, 2018 at 21:22
  • $\begingroup$ @NateEldredge: Indeed, it is a complex measure. $\endgroup$
    – Alex M.
    Jan 24, 2018 at 21:37

1 Answer 1

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It's easy to see that $||\mu||\le||\mu||_{TV}$. If $X$ is a locally compact Hausdorff space then the Riesz Representation Theorem says that $$||\mu||_{TV}=\sup_{f\in C_0(X),||f||=1 }|\int f\,d\mu|\le||\mu||.$$

I don't have a counterexample, but I tend to doubt that $||\mu||_{TV}\le||\mu||$ holds for a general topological space; you need some hypothesis on $X$ to ensure that there are "enough" continuous functions...

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  • $\begingroup$ On the other hand, the assumption that $\mu$ is regular excludes the usual counterexamples. $\endgroup$ Jan 24, 2018 at 21:14

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