# Why do we usually construct the Lebesgue Measure on finite outer measure sets before arbitrary measurable sets?

Question:

In the typical construction of the Lebesgue Measure in 6 stages (eg. in Lebesgue Integration on Euclidean Space by Jones), authors often construct the measure on sets that have finite outer measure before sets with arbitrary measure.

In Jones, he writes,

We should like to say that a set is measurable if $\lambda^* (A) = \lambda_* (A)$ (We say that such sets exist in $\mathcal{L}_0$). But if $A$ is so large that $\lambda_* (A) = \infty$, then this definition would not restrict $A$ at all.

Why is this step necessary? In the final definition of the Lebesgue measure, we have $\lambda(B) = \sup \{ \lambda ( B \cap A ) | A \in \mathcal{L}_0 \}$ where $B$ is a set with arbitrary measure, not necessarily finite outer measure. If it's possible that $\lambda(B) = \infty$, then shouldn't the definition of measurable in the quotation above suffice?

• If every set had measure $\infty$ this would still be a measure, but a silly one. You want to establish that finite measure sets actually exist. – Adam Hughes Jan 7 '15 at 2:54
• Surely the first method does not imply that every set had measure $\infty$, only those that have inner measure $\infty$ have measure $\infty$, which is what we want, correct? – mathjacks Jan 7 '15 at 11:21
• if I am not much mistaken, it does not rule it out, which is the problem. – Adam Hughes Jan 7 '15 at 11:22
• I'm still a little confused as to why that's an issue. Can you elaborate? – mathjacks Jan 7 '15 at 11:51