Let $K$ be a compact Hausdorff space. I want to show $\mathcal M(K)$ is an $\mathcal{l}_1$-sum of $L_1(\mu)$ spaces, where $\mathcal M(K)$ is the dual of $C(K)$. I have got the sketch of the proof but there are some small details that I don't know how to verify.

First, by Zorn's Lemma, we can choose a maximal collection of mutually singular probability measures, say $(\mu_i)_{i\in I}$ and claim that

$$\mathcal M(K) \cong \big(\oplus_{i\in I} L_1{(\mu_i)}\big)_{l_1}$$

For any $\nu \in \mathcal M(K)$ and $i \in I$, by Lebesgue's decomposition theorem, there exists finite signed measures $\lambda_i$ and $\rho_i$ such that $\nu=\lambda_i + \rho_i$, $\rho_i<<\mu_i $ and $\lambda_i \perp \mu_i$. Let $f_i := \frac{d\rho_i}{d\mu_i}$ be the Radon–Nikodym derivative.

Now my first question is:

How to show that $(f_i)_{i\in I} \in \big(\oplus_{i\in I} L_1{(\mu_i)}\big)_{l_1}$? By the Radon–Nikodym theorem, each $f_i \in L_1(\mu_i)$. But we need to show that the sum $\sum_{i\in I} \|f_i\|_1$ is finite. Do we actually have $\|v\|=\sum_{i\in I} \|f_i\|_1$?

After showing this, we define the map $\phi:\mathcal M(K) \to \big(\oplus_{i\in I} L_1{(\mu_i)}\big)_{l_1}$ by $\nu \mapsto (f_i)_{i\in I}$, which is an isometry. My second question is:

How to verify that this map is surjective?

Thank you in advance! (Sorry for the typos. I have corrected many of them.)


Given $\mu\in \mathcal{M}(K)$, note that for each finite set $J\subset I$ we have

$$\sum_{j\in J}\|f_j\|\leqslant \|v\|.$$


$$ \sum_{i\in I}\|f_i\|\leqslant \|v\|.$$


$$\nu_0 = \sum_{i\in I}f_i {\rm d}\mu_i.$$

Note that $\nu-\nu_0$ is singular with respect to all $\mu_i$ so by maximality, it must be the zero measure. Consequently,

$$\nu = \sum_{i\in I}f_i {\rm d}\mu_i$$

and $\|\nu\|=\sum_{i\in I}\|f_i \|$ so the map $\nu\mapsto (f_i)_{i\in I}$ is isometric.

As for surjectiviry, given $(f_i)_{i \in I}$, set $\nu(A) = \sum_{i\in I} \int_A f_i\,{\rm d}\mu_i$.

  • $\begingroup$ Thank you and +1! What about the surjectivity? $\endgroup$ – No One Nov 20 '15 at 2:56
  • $\begingroup$ @TiWen, I have added this. $\endgroup$ – Tomek Kania Nov 20 '15 at 15:43

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