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In this wikipedia article about Radon-measures, they made the following statement:

One way to do this is to define a measure on the Borel sets of the topological space. In general there are several problems with this: for example, such a measure may not have a well defined support. Another approach to measure theory is to restrict to locally compact Hausdorff spaces, and only consider the measures that correspond to positive linear functionals on the space of continuous functions with compact support.

Question: Why should such a measure not have a well defined support? And why is a well defined support important to the measure we want to define on the topological space?

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Suppose you have a set $S$ such that there is a probability measure $\mu$ defined on $(S,2^S)$ that assigns measure zero to every finite set. That such a set $S$ exists cannot be proven by the usual axioms of set theory, but nobody was successful showing it can't exist either. But then you have a Borel probability measure on $S$ with the discrete topology that fails to have a support.

If you are happy with a non-Hausdorff example, things are much easier. Let $U$ be any uncountable set and let a set be open if it is the empty set or has a finite complement. The Borel $\sigma$-algebra consists then of all sets that are countable or have a countable complement. The Borel probability measure that assigns measure $0$ to countable sets and measure $1$ to sets with countable complement has no support.

While working with a support is sometimes useful, I don't know any big theorems related to supports. On an intuitive level, the support tells us, modulo topological approximation, what area has positive measure.

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  • $\begingroup$ Just to be clear of notation: $2^S$ is the power set? $\endgroup$ – quallenjäger Jan 18 '18 at 23:01
  • $\begingroup$ @quallenjäger Yes. $\endgroup$ – Michael Greinecker Jan 18 '18 at 23:02
  • $\begingroup$ Thanks. Thats what I looked after. $\endgroup$ – quallenjäger Jan 18 '18 at 23:08

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