The dual of the space of continuous functions on a compact space $K$ is the space of regular signed measures with finite total variation, $rca(K)$. I have seen it stated (for example in this answer https://math.stackexchange.com/a/74877/564061) that this is isometric to $L_1(\mu)$ for some measure $\mu$ on $K$.

I know that the space is isomorphic to the $l_1$ sum of spaces $L_1(\mu_i)$ for mutually singular probability measures $\mu_i$, $i\in\mathcal{A}$ (possibly uncountable), see $\mathcal M(K)$ is an $\mathcal{l}_1-$sum of $L_1(\mu)$ spaces. Is there a way to relate this space to one of the form $L_1(\mu)$?

It seems to me that the obvious candidate bijection would map $(f_i)_{i\in\mathcal{A}}$ to $\sum_{i\in\mathcal{A}}f_i$. I think we can make sense of this sum since only countably many of the $f_i$ are non-zero at any $x\in K$, but I am not sure how to progress. At first I thought that for any $x$ there was at most one $i\in\mathcal{A}$ such that $f_i(x)\neq 0$, but that isn't the case.

I am asking because I wish to write elements of $C(K)^{**}$ as functions on $K$. The above relation allows us to equate $C(K)^{**}$ and the $l_{\infty}$-product of the spaces $L_{\infty}(\mu_i)$, and really I would just like a 'nice' relation between $C(K)^{**}$ and some space of functions, to avoid dealing with these nets of functions- the properties of the space aren't important. I am interested particularly in the case where $K\subseteq \mathbb{R}^2$.

  • $\begingroup$ The dual space of $C(K)$ is the space of finite Borel measures on $K$. $\endgroup$ – DisintegratingByParts May 22 '18 at 16:46
  • $\begingroup$ @DisintegratingByParts Yes I agree, but the dual of this space of measures is not nice. Really it is the bidual of $C(K)^{**}$ that I wish to consider. I thought that it may be possible to relate this space of measures to $L_1(\mu)$ and that would give some clearer representation of $C(K)^{**}$. $\endgroup$ – SEP101 May 22 '18 at 16:54
  • $\begingroup$ Your first paragraph states something about the dual of $C(K)$ that does not seem correct. Please check what you have there. $\endgroup$ – DisintegratingByParts May 22 '18 at 16:55
  • $\begingroup$ @DisintegratingByParts perhaps it is worded better now, or is there something still incorrect- perhaps my understanding is wrong. $\endgroup$ – SEP101 May 22 '18 at 17:03

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