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$\newcommand{\R}{\mathbf R}$ $\newcommand{\C}{\mathbf C}$

I am getting confused by the various different statements of the Riesz representation theorem found on the internet. So I want to clear up the confusion once and for all.

Let $X$ be a compact metric space.

$\bullet$ $M_s(X)$ denotes the set of all the finite signed Borel measures on $X$.

$\bullet$ $M_c(X)$ denotes the set of all the complex Borel measures on $X$.

$\bullet$ $M(X)$ denotes the set of all the finite (positive) Borel measures on $X$.

$\bullet$ $C(X, \R)$ denotes the space of all the real valued continuous maps on $X$ in the sup-norm topology. This is naturally a real linear space.

$\bullet$ $C(X, \C)$ denotes the space of all the complex values continuous maps in the sup-norm topology. This is naturally a complex linear space.

$\bullet$ $C(X, \R)^*$ denotes the space of all the real valued bounded real-linear maps with domain $C(X, \R)$.

$\bullet$ $C(X, \C)^*$ denotes the set of all the complex valued bounded complex-linear maps with $C(X, \C)$ as the domain.

A member $F\in C(X, \R)^*$ or $C(X, \C)^*$ is said to be positive if $F(f)\geq 0$ whenever $f\geq 0$.

RRT1. The map $M_s(X)\to C(X, \R)^*$ defined by $\mu\mapsto (f\mapsto\int_Xf\ d\mu)$ is bijective.

RRT2. The map $M(X)\to C(X, \R)^*$ defined by $\mu\mapsto (f\mapsto\int_Xf\ d\mu)$ is injective and has its image as the set of all the positive members of $C(X, \R)^*$.

RRT3 The map $M_c(X)\to C(X, \C)^*$ defined as $\mu\mapsto (f\mapsto \int_X f\ d\mu)$ is bijective.

RRT4 The map $M(X)\to C(X, \C)^*$ defined by $\mu\mapsto (f\mapsto\int_X f\ d\mu)$ is injective and has its image as the set of all the positive members of $C(X, \C)^*$.

Which of the above is/are true?

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If I am not mistaken all versions should be correct.

One way to proof RRT1 would be to first show RRT2 and then apply the Hahn-Jordan-decomposition for a given measure. (see Elias M. Stein and Rami Shakarchi. Functional Analysis: Introduction to Further Topics in Analysis, Chapter 1.7)

To verify RRT3 use RRT1 on the real and imaginary part of a given measure.

Finally the injectivity of RRT4 follows from RRT2. Let us assume that there is a positive functional $\varphi$, which is not the image of an element $m \in M(X)$. According to RRT3 we find a complex Borel measure $\mu + i\nu$ ($\mu,\nu \in M_s(X)$) so that $\varphi(f) = \int_Xf\ d(\mu + i\nu$). Assume that $\nu \neq 0$ then we find a Borel set $B$ with $\nu(B) \neq 0$. We can approximate the characteristic function $\chi_{B}$ via continuous positive functions $f_n$ in $L^1$ ($\rightarrow$ mollifiers). Therefore $\nu(B)= Im(\mu + i\nu(B)) = \lim_{n \to \infty} Im(\varphi(f_n) )=0$, since $\varphi$ is positive. Hence $\nu = 0$ and by RRT2 $\mu$ must be in $M(X)$. Thus RRT4 is true.

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    $\begingroup$ The map $\varphi$ doesn't seem to be complex linear. $\endgroup$ – caffeinemachine Sep 24 '18 at 4:38
  • $\begingroup$ That was my mistake. Sorry. I removed that part from the (partial) answer. $\endgroup$ – Richard Sep 24 '18 at 9:49
  • $\begingroup$ I think it should be correct now but I might be mistaken. $\endgroup$ – Richard Sep 24 '18 at 10:36
  • $\begingroup$ So I guess all four are correct. Thanks. Your answer helps. $\endgroup$ – caffeinemachine Sep 24 '18 at 16:17

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