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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?

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You picked the wrong direction to disprove. Being measurable implies the equality of inner and outer measures under any circumstances. Try a counterexample in the opposite direction. Hint: $1+\infty=2+\infty$. – user31373 May 22 '12 at 20:43
Oh, I think I've got it then. Union the Vitali set with the negative reals and then the outer measure is $\infty$ and the inner measure is $\infty$. But the set is unmeasurable, since its invariant under rational translations. Does that make sense? – Sam Kidd May 22 '12 at 22:32
The union (call it $E$) is not invariant under rational translations.... Instead, recall that the intersection of two measurable sets is also measurable. So, if you can find a measurable set $B$ such that $E\cap B$ is known to be not measurable, you are done. – user31373 May 22 '12 at 22:44
$E \cap [0,1]$ isn't measurable because it is the Vitali set, even though $[0,1]$ is measurable. Therefore $E$ isn't measurable, even though $m^*(E) = m_*(E) = \infty$. – Sam Kidd May 22 '12 at 22:49
up vote 3 down vote accepted

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.

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Theorem: We know that $m_*(E) \leq m^*(E)$, and that equality holds if and only if $E$ is Lebesgue measurable.

Now any non-measurable set will serve our purpose.

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I think you have missed word "measurable" in the question title. – Wojowu Aug 16 '15 at 8:43

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