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. 



Borel sets and Method I measure
Anything degenerate enough will do.


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 $\mu(A\cap E)=\mu(A\cap(X\setminus E))=\mu(A)=1$.


