Linear forms on the algebra of functions

Question

In the book "Information Geometry", Nihat Ay et al. consider the following scenario in Chapter $2.1$ on page $24$:

We consider a non-empty and finite set $I$. The real algebra of functions $I\rightarrow \mathbb{R}$ is denoted by $\mathcal F(I)$, and its unity $\mathbb 1_{I}$ or simply $\mathbb 1$ is given by $\mathbb 1(i) = 1$, $i\in I$. [...] We naturally interpret linear forms $\sigma:\ \mathcal F(I) \rightarrow \mathbb R$ as signed measures on $I$ and denote the corresponding dual space $\mathcal F(I)^{\star}$ [...].

My problem is understanding how linear forms $\sigma$ can be interpreted as signed measures on $I$? After all, $\mathcal F(I)$ is not a sigma-algebra, since it is not even a subset of the power set $\mathcal P(I)$?

Answer 1

I see how you got confused here, since $\sigma : \mathcal{F}(I) \to \mathbb{R}$ looks like it might itself describe a measure. After all, some people use $\mathcal{F}$ as notation for a $\sigma$-algebra, and signed measures should be functions from a $\sigma$-algebra to $\mathbb{R}$. The authors should probably have been more careful, since this is not at all what's meant!

Instead, $\mathcal{F}(I)$ should be thought of as continuous functions $I \to \mathbb{R}$, where $I$ is given the discrete topology. In this sense, maybe a better name for $\mathcal{F}(I)$ would be $C(I,\mathbb{R})$. Of course, the (measure theoretic) Riesz Representation Theorem tells us that

For every linear functional $\varphi : C(I,\mathbb{R}) \to \mathbb{R}$, there is a unique (signed, radon) measure $\mu_\varphi$ on $I$ so that $\varphi(f) = \int_I f \ d\mu_\varphi$

This $\mu_\varphi$ is slightly tricky to define in general, but since $I$ is a discrete space it's really not hard at all! Here $\mu_\varphi(E)$ is defined to be $\varphi( \chi_E )$, where $\chi_E : C(I, \mathbb{R})$ is the characteristic function of $E \subseteq I$.

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I hope this helps ^_^