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Let $X$ be a compact, Hausdorff space. Let $B(X)$ be the Banach space of bounded, Borel-measurable, complex-valued functions on $X$ under the uniform norm. Let $C(X) \subset B(X)$ be the closed subspace of continuous, complex-valued functions.

By the Riesz representation theorem, we have an isomorphism $M(X) \cong C(X)^*$ where $M(X)$ denotes the (finite) regular, complex Borel measures on $X$ under the total variation norm. The isomorphism sends $\mu \in M(X)$ to integration against $\mu$. By the Hahn-Banach theorem, the element of $C(X)^*$ corresponding to $\mu \in M(X)$ extends to an element of $B(X)^*$ with the same norm. In fact, since the functions in $B(X)$ can also be integrated against $\mu$, we have a rather canonical choice of extension. It is not difficult to see that $f \mapsto \int_X f \ d\mu$ is in $B(X)^*$ and has the same norm as its restriction to $C(X)$.

Question: Is it true that every functional in $C(X)^*$ has a unique norm-preserving extension to $B(X)^*$ (given by integration against the corresponding measure)? Or, if not, what sort of functional analysis-type statements can be made in order to single out this extension which is obviously the preferred one from a measure theory standpoint?

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From Hildebrandt-Knatorovich theorem we know that $B(X)^*$ is isometrically isomorphic to the Banach space of finitely additive (not necesarily regular) complex valued measures of bounded variation. This space is denoted by $ba(X)$. Recall also that $C(X)^*$ is $M(X)$. Now the mapping $$ i: M(X)\to B(X)^*, \mu\mapsto \left(f\mapsto\int_X fd \mu\right) $$ is nothing more than the natural embedding of the space of good measures $M(X)$ into the space of weird measures $ba(X)$. It is worth to say that $M(X)$ is nicely situated in $ba(X)$ - it is norm one complemented via projector $j_C^*:ba(X)\to M(X)$. Here $j_C:C(X)\to B(X)$ is a natural embedding.

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  • $\begingroup$ This does not seem to answer the question. The OP asks if what you denote by $i$ is the only way to "promote" objects from $M(X)$ to objects in $ba(X)$, not about its properties. $\endgroup$ – Alex M. Mar 30 '18 at 19:44

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