Riesz-Representation-Theorem states that every positive linear functional $F$ for any finite continuous $f$ on a local compact space S one can find a unique borel measure, such that $$F(f)=\int f d\mu$$.
- In the dual space there are not only positive linear functional. The dual space includes all linear functional. So according to the version of Riesz-representation theorem that I stated above, the set of finite borel-measure is only a subset of the dual space. What is the sufficient condition, that every element in the dual space can be written in the form of Riesz-Representation? Will the continuity of the functional help if we introduce the weak topology on the space?
- What about if I take a separable metrizable space S instead of local compact space S? Can we still apply the theorem?