Constructing the Lebesgue measure on the reals.

Question

So far, we have only constructed the Lebesgue measure on intervals. Now if $\{I_k\}$ is a countable cover of the reals by intervals which overlap at one point at most, we say that $S\subset \mathbb R$ is measurable iff $S \cap I_k $ is measurable in $I_k$ for all $k$ and set $$m(S)=\sum_{k=0}^{\infty}m_{I_k}(S\cap I_k)$$ I have shown that the values coincide for any two covers, knowing that $S$ is measurable for all elements in both covers. How do we show that if S is measurable according to one cover, then it is measurable according to another? (Then showing that the family of measurable sets is a $\sigma$-algebra and additivity of the measure should pose little problem.)

Answer 1

You should show that if $I\subset J$ and $S\cap I$ is measurable in $I$ then $S\cap I$ is measurable in $J$. Then given any two covers you can reduce them to a common subcover.