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I've seen a similar question on this forum before, yet i could not find a solution to this particular problem. Could you give me some tips on how to proceed with the solution? Here's the statement of the problem: Let $A$ be a bounded set in $\mathbb R^k$. Let $K$ be a $G_\delta$ set containing $A$, such that the outer measure of $A$ is equal the measure of $K$. Prove that $A$ is measurable.

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Since you are a new user, here are some tips. for better results try showing some work or explaining the steps you tried and where you got stuck. Also make sure to remember to accept an answer by clicking the check mark next to answers provided by other users, once you find one which satisfies your question. – MSEoris Dec 17 '12 at 19:18
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For any set $A \subset \mathbb R^n$ you can find a $G_\delta$ subset containing $A$ of the same outer measure. This follows because you can always find an open set $U$ containing $A$ such that $m^\ast(U)<m^\ast(A)+\varepsilon$, this is by the definition of outer measure. So by taking appropriate intersections you can find such a $G_\delta$ set. In particular your question is false because there exist non-measurable subsets.

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Oh yes, i feel plain stupid not having thought about it. I could prove all you wrote in the first 3 sentences yet i tried hopelessly to reverse the situation not considering non-measureable sets. Thank you so much for the insight. – mateusz Dec 17 '12 at 19:13

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