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I was thinking if it is possible to come up with a $\sigma$-finite measure on $\mathbb R$ which is positive on any uncountable set. I think that I have a proof that there is no such measure - but I am not sure if it is formal enough.

The proof: let $\mu$ be such measure, then for any $[a,b]$ such that $a it holds that $\mu([a,b]) = \infty$. To show it, we consider a bijection $f:[a,b]\to K$ where $K = [0,1]\times[0,1]$ and put $ f_x:=f^{-1}([0,1]\times\{x\})\subset[a,b] $ for all $x\in [0,1]$. Clearly, $f_x$ is uncountable and hence $\mu(f_x)>0$. Now, if there is only finitely many $x$ such that $\mu(f_x)>1/n$ for all $n\in\mathbb N$ then we obtain that there are only countably many $x$. As a result, for some $n\in\mathbb N$ there are infinitely many $x$ such that $\mu(f_x)>1/n$ and hence $\mu([a,b])=\infty$.

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    @N.S.: $C-C=[-1,1]$, so $C\cap(C+t)\ne\varnothing$ for $t\in[-1,1]$.2012-02-14

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You almost got it, your argument actually shows that for any uncountable $A\subseteq \mathbb R, \mu(A)=\infty$ (just split $A$ into uncountable many uncountable pieces). Now, if $\mathbb R=\bigcup_{n=1}^\infty A_n$ then some $A_n$ must be uncountable so $\mu(A_n)=\infty$. Therefore $\mu$ is not $\sigma$-finite.

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    Yes, that should work. I don't know how complicated it is to prove the result azarel mentioned. But if you use the Borel isomorphism theorem anyways (which is hard to prove), you can split any uncountable Borel set into $\mathfrak{c}$ Borel sets in essentially the way you used: Every uncountable Borel set is isomorphic to $[0,1]^2$, so you can split it up into the isomorphic copies of the slices $[0,1]\times\{x\}$ for $x\in[0,1]$.2012-02-15