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This is the definition in Stein's Real analysis.

For any set $E\subset X$, outer measure of $E$ is defined as $$ m_{*}\left(E\right)=\inf\sum_{i=1}^{\infty}Q_{i} $$ where infimum is taken over all closed balls $Q_{i}$ satisfying $E\subset\bigcup_{i=1}^{\infty}Q_{i}$.

Moreover, If $E$ is a Lebesgue measurable set, then the Lebesgue measure on $E$ is defined $$ m(E) = m_{*}\left(E\right). $$

Here is my question. Let finite measurable set $E$ be given.

Then some people use the following argument : For given $\epsilon>0$, there exists closed sets $\left\{ Q_{i}\right\} _{i=1}^{N}$ such that $\bigcup_{i=1}^{N}Q_{i}$ contains $E$ and $m(\bigcup_{i=1}^{N}Q_{i}\setminus E)<\epsilon$.

In my opinion, this is wrong. It should be following : For given $\epsilon>0$, there exists closed balls $\left\{ Q_{i}\right\} _{i=1}^{\infty}$ such that $\bigcup_{i=1}^{\infty}Q_{i}$ contains $E$ and $m(\bigcup_{i=1}^{\infty}Q_{i}\setminus E)<\epsilon$.

Am I being wrong? Or can we use both statement? Is the above statement equivalent? Thanks in advance.

kayak
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  • No, it is in fact correct. I think that there is even a stronger statement that can be made. No proof at the moment, though. –  Jan 03 '17 at 15:01
  • @OpenBall Why is there no proof at this moment? If you can, please reply to this question for me. – kayak Jan 04 '17 at 03:33
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    Do you really mean $E$ to be finite, or to have finite measure? If the latter then the "some people" argument is wrong. For a simple example take $E = \mathbb{Q}$. If $Q_i$ are closed and $\bigcup Q_i$ contains $E$ then $\bigcup Q_i = \mathbb{R}$ and we have $m(\bigcup Q_i \setminus E) = \infty$. – Nate Eldredge Jan 04 '17 at 05:08
  • @NateEldredge Thanks for comment. (it is finite measure)Your comment reminds me of my previous question : http://math.stackexchange.com/questions/2070246/rational-numbers-are-measure-zero. You mean if we cover $\mathbb{Q}$ by closed sets, the difference of two sets has infinite measure. In my link, the 'open' and 'closed' are different? In my opinion, the answer on my link could be eligible to closed sets. – kayak Jan 04 '17 at 05:16
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    I am not sure I understand your comment. But if you take the "some people argument" and replace "closed" by "open" then it is true. – Nate Eldredge Jan 04 '17 at 05:26
  • @NateEldredge Okay. How about finite closed sets? "Some people argument" has its cover the finite number of closed sets. – kayak Jan 04 '17 at 05:29
  • It doesn't help. A finite union of closed sets is still a closed set, so if you can't get a good cover with one closed set, you can't get it with a finite union of closed sets either. – Nate Eldredge Jan 04 '17 at 05:33
  • @NateEldredge Thanks. I understand your comment. But now, I'm confused on why 'my argument' is true. It has union of closed sets. Then is my argument is also false? – kayak Jan 04 '17 at 05:58
  • Your argument allows for infinite unions of closed sets. An infinite union of closed sets is not necessarily closed. Your claimed statement is true. – Nate Eldredge Jan 04 '17 at 06:02
  • @NateEldredge Getting clear. How about changing '$m$' to outer measure '$m_{*}$' on 'some people' argument and any subset(including non-measurable sets) in $X$? Is it then true? – kayak Jan 04 '17 at 06:08
  • No, you just made that claim stronger. It's still false and the counterexample I gave above is still a counterexample. – Nate Eldredge Jan 04 '17 at 06:10
  • @NateEldredge Ah!! So thank you! I just remember it is the $F_{\sigma}$-set and the definition of outer regular and inner regular. Happy New Year. What should I do to this question? Should I remove? I don't to do that. – kayak Jan 04 '17 at 06:16
  • Sorry, I must have misread the question (there's a closely related result). –  Jan 04 '17 at 12:47

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