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This question was asked here, but the question wasn't asking for an answer, just a clarification on the proof by contradiction they already had, so I'm reposting it to ask for help arriving at the answer.

Because the asker above said they proved it by contradiction, that is what I'm attempting. But I'm having issues wrapping my brain around the concept of set measurability. So let's give it a good college try with what I can understand.

For clarity, $\mu$ is an outer measure.

A is nonmeasurable, so $\exists B \subseteq X$ such that $\mu (B) \ne \mu (B \bigcap A) + (B \bigcap A^c)$.

E is measurable, so $\mu (B) = \mu (B \bigcap E) + (B \bigcap E^c)$.

This essentially means $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$ (Since monotonicity is alredy implied by the definition of outer measure).

So we can attempt a proof (or the start of one) by contradiction.

Assume $\mu (E \setminus A)=0$. So $E \setminus A$ is a null set.

So $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$. I assume I'm supposed to show a contradiction using that A is nonmeasurable. I'm unsure how.

Another way I think could work would be to use: $\mu (E \setminus A)= \mu (E \bigcap A^c) = 0$ and arrive at a contradiction given that A is nonmeasurable, but again, I'm unsure how.

I just need some direction from here. Thank you in advance for any advice.

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Measures induced by outer measures are complete, meaning that any subset of $X$ with outer measure zero is measurable. To see this, suppose that $S \subseteq X$ and $\mu(S) = 0$. Let $Y \subseteq X$ be any subset. By monotonicity, we have $\mu(Y \cap S) \leq \mu(S) = 0$, so $\mu(Y \cap S) = 0$. Also by monotonicity, we have $\mu(Y \cap S^c) \leq \mu(Y)$.

Therefore, $$\mu(Y \cap S) + \mu(Y \cap S^c) = \mu(Y \cap S^c) \leq \mu(Y)$$ which shows that $S$ is measurable.

Now, let $E$ and $A$ be as in your problem. Suppose for a contradiction that $\mu(E \cap A^c) = 0$. Then by the above argument, $E \cap A^c$ is measurable. But $E$ is also measurable, so this means that $A = E \setminus (E \cap A^c)$ must be measurable. But this is a contradiction.

So, our assumption that $\mu(E \cap A^c) = 0$ is untenable; hence we must have $\mu(E \cap A^c) > 0$.

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  • $\begingroup$ I hope to one day make a proof so elegant. Thank you so much! $\endgroup$ – Lo12 Nov 2 '17 at 17:27

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