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Let $X$ be a non-empty set, let $\mathcal{A}$ be an algebra of sets on $X$, and let $\mu$ be a finite content on $(X,\mathcal{A})$. Can $\mu$ be extended to a content on $2^X$ (i.e. on $X$'s power-set)?

* As far as I know, the Lebesgue measure on $\mathbb{R}$ (and hence the Jordan content on $\mathbb{R}$) can be extended to a content on $2^{\mathbb{R}}$, but can this be done in general?

Asaf Karagila
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Evan Aad
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1 Answers1

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Let $\ell^\infty(X)$ be the Banach space of bounded functions on $X$ equipped with the $\sup$ norm. Let $Y$ be the subspace spanned by the set $\{ \chi_A : A \in \mathcal{A} \}$ where $\chi_A$ is the indicator function of $A$.

Notice that a finite content $\mu$ on $(X, \mathcal{A})$ defines a linear functional on $Y$ by integration, which I will denote by $\phi$. That is, for $f \in Y$, $\phi(f) = \int f d \mu$. In particular, this gives us that $\phi$ is bounded since the usual properties of the Lebesgue integral for simple functions are true for integration of simple functions against a finite content.

By the Hahn-Banach Theorem, we can extend $\phi$ to a bounded linear functional $\tilde{\phi}$ on all of $\ell^\infty(X)$. You can then check that $\tilde{\mu}(A) = \tilde{\phi}(\chi_A)$ defines a finite content which extends $\mu$ to the power set of $X$.

Rhys Steele
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  • Thanks. 1. What is $\mathcal{A}$? 2. What is this example meant to show? It doesn't seem to be a counter-example to the claim stated in my question, but it doesn't seem to prove it either. – Evan Aad May 07 '18 at 16:23
  • $\mathcal{A}$ is the same thing it is in your question? This answer shows the existence of an extension of an arbitrary finite content $\mu$ on $\mathcal{A}$ which is $\tilde{\mu}$ in my notation – Rhys Steele May 07 '18 at 16:26
  • I see. Thanks. When you say that $\mu$ defines a linear functional on $Y$ by integration, what kind of integration do you have in mind? Lebesgue integration is defined w.r.t. a measure, not a content. – Evan Aad May 07 '18 at 16:30
  • Here you don't even need a proper integration theory since I only need to define $\phi$ on the space $Y$ of simple functions for the algebra $\mathcal{A}$. So for this, potentially the simplest thing to do is use the usual Lebesgue construction which will work for our purposes without the assumption of countable additivity. – Rhys Steele May 07 '18 at 16:36
  • I see. In order to use the Hahn-Banach theorem, the linear functional determined by $\mu$ via integration needs to be dominated by a sublinear functional that is defined on the entire space $\ell^{\infty}(X)$. What is this sublinear functional in your example? – Evan Aad May 07 '18 at 16:50
  • Sorry, I didn't realise the link I'd provided went to the sublinear functional version of Hahn-Banach. There is a simpler, special case that says that a bounded, linear functional defined on a subspace of a normed vector space can be extended to a bounded linear functional of the same norm on the whole space (take the sublinear functional to be a suitable scalar multiple of the norm). Here the sublinear functional is $\mu(X) | \cdot |_\infty$. – Rhys Steele May 07 '18 at 16:56
  • Let us continue this discussion in chat. – Rhys Steele May 07 '18 at 16:58
  • I'm finding the use of the word "integration" to be a bit opaque here. What you mean is $\phi(\chi_{A}) = \mu(A)$ and then extend linearly to $Y$, is that right? It might help to clarify this in your post. –  May 07 '18 at 18:18
  • @fourierwho Yes, that is a rephrasing of the definition of the Lebesgue integral which works here... – Rhys Steele May 07 '18 at 19:17