$A$ uncountable thus $\mu(A)>0$

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 $a<b$ 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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If $\mathbb R=\bigcup_n A_n$ (with $A_n$ of finite measure) then at least one $A_n$ is uncountable, otherwise $\mathbb R$ would we be countable. It's a contradiction. – Davide Giraudo Feb 14 '12 at 21:23
What is your sigma algebra? Most uncountable sets are not Borel measurable... if you use the Borel-sigma algebra, then $f_x$ is probably not measurable... If you use the parts of $R$ as a sigma algebra, it is easy to partition $R$ into infinitely many translates of the same uncountable set.... – N. S. Feb 14 '12 at 21:24
@N.S. thanks for the comment. I think I am interested in the case when sigma-algebra is Borel. – Ilya Feb 14 '12 at 21:32
@Ilya: I wonder if one could embed infinitely many different translates of the Cantor set into a finite interval. That would do it, but it is not clear if it is possible or not.. Since the Cantor set has Lesbegue measure zero, I don't see any obvious reason why that couldn't be done, but it doesn't mean it can.... – N. S. Feb 14 '12 at 21:37
@N.S. That was my main doubt: is it possible to divide an uncountable set into uncountable many uncountable sets, and if yes - could they be Borel measurable (provided the original set is) – Ilya Feb 14 '12 at 21:40

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.
$\mathbb R$ can be partitioned in $\aleph_1$ Borel sets. I don't know if this is possible for all borel sets though. – azarel Feb 14 '12 at 22:18
@MichaelGreinecker so the complete argument would be: suppose that there is $\sigma$-finite measure $\mu$ which in positive on any uncountable Borel susbet of $\mathbb R$. Then any uncountable Borel set has an infinite measure since it can be partitioned with an uncountable number of uncountable Borel sets (here your comment is used). Then $\mathbb R = \bigcup_n A_n$ and hence at least one $A_n$ is uncountable, so $\mu(A_n) = \infty$ and we obtain the contradiction. Is it right? – Ilya Feb 15 '12 at 8:49
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 $x\in[0,1]$. – Michael Greinecker Feb 15 '12 at 9:15