# Borel sets and Method I measure

If $$(X,d)$$ is a metric space, say a function $$\tau$$ on some class $$\mathscr{C}$$ of subsets of $$X$$ is a pre-measure, if $$\emptyset \in \mathscr{C}$$, $$\tau(\emptyset)=0$$ and $$0\le \tau(C)\le +\infty$$ for all $$C\in \mathscr{C}.$$

If $$\tau$$ is a pre-measure on some class $$\mathscr{C}$$ of subsets of $$X$$, then the set function $$\mu(E)=\inf\left\{\sum_{i=1}^\infty\tau(U_i):U_i\in \mathscr{C}, E\subset \cup_{i=1}^\infty U_i\right\}$$ is a measure on $$X.$$ The measure is called Method I measure. For more detail, see C.A. Rogers's book titled Hausdorff measure.

One claims that "Borel sets are not in general measurable with respect to measures from Method I constructions" in a literature. Can someone give an example? Thanks.

Suppose $$\mathcal C=\{\emptyset,X\}$$ and, say, $$\tau(X)=1$$. Then $$\mu(E)=1$$ for every non-empty $$E\subset X$$. Therefore any set $$E\subset X$$ which is not $$X$$ nor $$\emptyset$$ will not be measurable. Indeed take $$A=\{x,y\}$$ where $$x\in E$$ and $$y\in X\setminus E$$. Then $$\mu(A\cap E)=\mu(A\cap(X\setminus E))=\mu(A)=1$$.