I'm working on providing a counterexample to the claim that
A unbounded set $A \subset \mathbb{R}$ is Lebesgue measurable if and only if its inner and
outer measures are equal. Further, if $B$ is an unbounded measurable set that contains $A$,
then $A$ is measurable if and only if it divides $B$ cleanly.
Let me clarify which definitions I'm using.
Lebesgue outer measure is
$$m^*A = \inf\left\{ \sum_k |I_k| : \{I_k\} \text{ is a covering of $A$ by open intervals}\right\}$$Lebesgue inner measure is
$$m_*A = \sup\left\{ m^*C : C\text{ is closed and }C \subset A\right\}$$A set $E$ is Lebesgue measurable if the division $E|E^c$ of $\mathbb{R}$ is so "clean"
that for each "test set" $X \subset \mathbb{R}$, we have
$$m^*X = m^*(X \cap E)+ m^*(X \cap E^c)$$
So far, I have thought that the best strategy to disprove the claim is to find an example of an unbounded measurable set that has unequal inner and outer measure. Does this seem like the right direction? Are there any example sets I should study?
Best Answer
The counterexample is as follows. First, we define an unmeasurable set: $V \subset [0,1]$ (perhaps something like the Vitali set). Then, defining $R = (2,\infty)$, we can consider our counterexample function $C = V \sqcup R$. We know that $C$ is unmeasurable because if it were, since $[0,1]$ is measurable, we know that $C \cap [0,1]$ would be measurable. This clearly isn't the case because $C \cap [0,1] =V$.
Now, we know that $m^*C = m_*C = \infty$, because $m^*R = \infty$ and both inner and outer measure are monotonic. Therefore, we have an unbounded, unmeasurable set with equal inner and outer measure, contradicting the claim.