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Shows that all compact of $\mathbb R$ is the support of a Borel measure.

Comment. For a measure $\mu$ I have to shows that

$supp(\mu)$ = $K$.

If $\mu(X) = 0$, then all compact satisfy $\mu(K) = 0$.

If $\mu(X)< +\infty$ then exists $M \in \mathbb R$ that $\mu(X) = M$, next

supp $\mu$ = $Cl{[\mu \neq 0 ]} = Cl\{ x \in X, \mu(x) > 0, \mu(x)\leq M\} = Cl( \mu^{-1}(0,M])$. Where $( \mu^{-1}(0,M])$ is a borel set.

Next, inside that open set we can find a compact K and then $ K \subset supp(\mu) $.

On other hand, $X = \bigcup^{\infty} K_n$ and then $ supp(\mu) \subset \bigcup^{\infty} K_n$.

Well this is that I think, is convincing?

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I'm afraid that this is not convincing at all. You are supposed to construct a measure $\mu$ which is supported in $K$. Where is this measure? – Giuseppe Negro Jan 27 '13 at 3:26
Think of counting measure. – Jose27 Jan 27 '13 at 4:58
I'm still thinking, following the suggestion. Let $\mu$ counting measure. If $\mu(A) = \infty$ then $A$ is infinity then take $K = \bigcup K_{n}$ we have $supp(\mu) \subset K$. If $\mu(A)< \infty$ then $A$ is finite, then take each point of $A$ and let $K = \bigcup_{a\in A} \{a\}$ is a compact, then $supp(A) = K$. – Malaq Jan 27 '13 at 19:28
I really don't understand what you're doing, what does $A$ have to do with this? – Jose27 Jan 27 '13 at 20:58
What did you mean with counting measure, I don't understand. I know the definition of this, but, how to work with it? – Malaq Jan 27 '13 at 21:31

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