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.

  • $\begingroup$ 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. $\endgroup$ – user384138 Jan 3 '17 at 15:01
  • $\begingroup$ @OpenBall Why is there no proof at this moment? If you can, please reply to this question for me. $\endgroup$ – kayak Jan 4 '17 at 3:33
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    $\begingroup$ 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 $m(\bigcup Q_i \setminus E) = \infty$. $\endgroup$ – Nate Eldredge Jan 4 '17 at 5:08
  • $\begingroup$ @NateEldredge Thanks for comment. (it is finite measure)Your comment reminds me of my previous question : math.stackexchange.com/questions/2070246/…. 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. $\endgroup$ – kayak Jan 4 '17 at 5:16
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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Jan 4 '17 at 5:26

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