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I fear this question has already been asked but I couldn't find anything with the search tool.

Our definition of the Lebesgue outer measure on $\mathbb{R}^n$ is $$ \nu (A)= \inf \{ \sum_{I=1}^\infty Vol(Q_i) | Q_i \in \mathcal{Q}_n ,A\subset \bigcup_i Q_i \}$$ where $\mathcal{Q}_n$ is the set of all $n$-dimensional cuboids, i.e. sets of the form $Q_i=[a_1,b_1] \times \dotsb \times [a_n,b_n]$ with $(a_1,...,a_n),(b_1,...b_n) \in \mathbb{R}^n$

Would there be any problem if we only allowed finite sums in the definition?

The Lion King
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    I don't quite know if cuboids are products of bounded intervals or of intervals, period, but either way the answer is yes: for instance you would have $\nu(A)=\infty$ for all unbounded open sets. – Sassatelli Giulio Jan 27 '23 at 09:06
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    How would you propose that we get the measure under the graph of $1/x^2$ on $[1,\infty)$? Such sets that might have a nicely defined measure despite extending out infinitely in one direction get left out. (I assume your cuboids are themselves finite in measure / bounded.) – PrincessEev Jan 27 '23 at 09:09
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    See "Jordan measure." (A bounded example: the Jordan measure of $[0,1]\cap\mathbb{Q}$ in $\mathbb{R}$ is $1$, but its Lebesgue measure is $0$.) – Noah Schweber Jan 27 '23 at 09:21

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