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The addition property of double integrates states the follows : enter image description here

On the left, We are calculating over the region twice and this it makes to subtract the integral over that line that is common to them (i.e, their common boundary) from the right. However, the only reasoning I could find was either simply stating that the boundary is a line and thus has no area or using measure theory such as here.

Can we rigorous show the above formula without resorting to measure theory?

watcher54
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    Sure you can. Just show that the intersection is a set of measure zero in the non-measure theoretic sense. I.e. for every $\varepsilon>0$, there exists a finite union of open boxes $U$, such that the area of $U$ is at most $\varepsilon$ and use this to prove that $|\int \int_{S\cap T} fdA|\leq \int \int_U |f| dA\leq |f|_{\infty} \varepsilon$. Of course, for the existence of $U$ to be nice to prove, you probably have assume some amount of regularity on $S\cap T$ (i.e. that it is piecewise $C^1$ or something like that). – WoolierThanThou Jun 17 '21 at 07:22
  • Thank you. What does $ | f |_{\infty} \varepsilon$ mean? – watcher54 Jun 17 '21 at 07:37
  • $|f |_{\infty}$ is the supremum of $|f|$. $\varepsilon$ is the $\varepsilon$ from before. – WoolierThanThou Jun 17 '21 at 07:42
  • thank you @WoolierThanThou. – watcher54 Jun 17 '21 at 07:56

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