extension of a non-finite measure For a finite measure on a field $\mathcal{F_0}$ there always exists its extension to $\sigma(\mathcal{F_0})$. 
Can somebody give me an example of a non-finite measure on a field which cannot be extended to $\sigma(\mathcal{F_0})$.
It would be better if somebody can point in the proof (for existence of extension of finite measure), where the finite-ness property is used.
 A: Every measure can be extended from a field to the generated $\sigma$-algebra. The classical proof by Caratheodory does not rely on the measure being finite, so there is no such example. As Ilya mentioned in a comment, the extension may not be unique. Here is an explicit example:
Let $\mathcal{F}$ be the field of subsets of $\mathbb{Q}$ generated by sets of the form $(a,b]\cap\mathbb{Q}$ with $a,b\in\mathbb{Q}$. Let $\mu(A)=\infty$ for $A\in\mathcal{F}\backslash\{\emptyset\}$. It is easy to see that $\sigma(\mathcal{F})=P(\mathbb{Q})$, the powerset. Now let $r>0$ be a real number. Then there is a unique measure $\mu_r$ such that $\mu\big(\{q\}\big)=r$ for all $q\in\mathbb{Q}$. So for each $r>0$, $\mu_r$ is a different extension of $\mu$. So there is a continuum of possible extensions of $\mu$ to $\sigma(\mathcal{F})$. The example is based on one in Counterexamples in Probability and Real Analysis by Wise and Hall.  
A: Since Carathéodory's extension theorem does not require that the spaces are sigma-finite, there is no such counterexample. (As already mentioned, in the absence of sigma-finiteness, the uniqueness disappears, not the existence.)
