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We have defined the outer measure (in $\Bbb{R}^n$) in class as:

$$\mu (A) = \inf \left\{ \sum_{i=1}^\infty v(I_i)\quad \{I_i\}_{i=1}^\infty \text{ collection of open cubes with } A \subset \bigcup_{i=1}^\infty I_i \right\}.$$

I have proved, as our exercises suggested, that the definition holds if we use $I_i$ as open and bounded cubes and also if $I_i$ are compact cubes.

However, since the exercise doesn't suggest to do the same with closed (not necessarily bounded) cubes, I suppose it's false.

Is this true? Would you know any counterexample?

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    $\begingroup$ It does not matter whether you use open/closed/compact cubes/rectangles/balls, etc. $\endgroup$ – Sangchul Lee Oct 24 '18 at 8:10
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    $\begingroup$ If you try and think of a "new" form of approximation, like closed cubes, you should ask yourself - don't these have open cubes inside of them? Didn't I already show that it works in this case? $\endgroup$ – Prototank Oct 24 '18 at 13:01

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