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The original riesz representation theorem states

Let $X$ be locally compact hausdorff space. Then for any nonnegative functional $\Lambda$ on $C_c(X)$, there is a unique regular borel measure $\mu$ on $X$ such that $$\Lambda(f)=\int f\mu(dx)$$ for all $f\in C_c(X)$.

Suppose there is a measured space $(\Omega,\mathcal{A},P)$. I define the space of measurable functions $L^0(P)$ is the equivalence class of measurable functions, which are $P$-a.s. equal. This is a vector space and often used in mathematical finance. If we suppose that $\Omega$ is finite, then we identify $L^0$ with $\mathbb{R}^n$ (for $Y\in L^0$, $Y(\omega_i):=y_i)$.

I was able to prove that a given functional $\Gamma$ on $L^0$ is nonnegative. Can I use the Riesz-Representation theorem to conclude that $$\Gamma(Y)=\int Y R(dw)$$ for a measure $R$?

The problem is the hyptothesis about $C_c(X)$. My notes say there is such a measure, but I'm not sure, why I can apply Riesz.

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If $\Omega$ is finite, meaning that it is a finite set (which is what you use to identify $L^0$ with $\mathbb{R}^n$), then you do not need anything of this high-powered machinery. No locally compact hausdorff, no compactly supported functions, no topology at all. – Giuseppe Negro Jun 20 '13 at 12:41
Indeed in this case you have that $\Gamma$ is essentially a row vector with positive entries and "integration against the measure $R$" reduces to a dot product against this row vector. This language makes everything much simpler. – Giuseppe Negro Jun 20 '13 at 12:43
@GiuseppeNegro stupid me! You can turn your comment into an answer, then I will accept it. – math Jun 20 '13 at 12:51
Ok, I'll do it. – Giuseppe Negro Jun 20 '13 at 12:52
up vote 4 down vote accepted

If $\Omega$ is finite then $\Gamma$ is essentially a row vector with positive entries and "integration against the measure $R$" reduces to a dot product against this vector.

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As Giuseppe Negro said this is overkill. A linear functional on $L^0 = \mathbb{R^d}$ is indeed a represented as a vector (any linear function between finite dimensional vector spaces can be represented a matrix, if the arrival space is $\mathbb{R}$ then it is a vector). The measure in this case is calculable as $$ R_k = \Gamma(e_k) $$ where $e_k$ is the $k$-th basis vector (normally the canonical base). And so your functional is written as $$\Gamma(Y) = \Gamma((Y_1, \ldots, Y_d)) = \sum_{i=1}^d Y_i R_i = \int_{\{1, \ldots, d\}} Y dR $$ thinking as $Y \in \mathbb{R}^d$ as a function from $\{1, \ldots, d\}$ to $\mathbb{R}$.

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