# What does the Riesz representation theorem say?


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?

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.
• The map $\varphi$ doesn't seem to be complex linear. – caffeinemachine Sep 24 '18 at 4:38