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Here is the full problem:

Suppose $E_1,E_2 \subseteq \mathbb{R}$ are measurable sets. Show that $$m(E_1 \cup E_2) + m(E_1\cap E_2)=m(E_1)+m(E_2)$$

My attempt has been as such:

Suppose $E_1, E_2$ are measurable. Now, recall that $$E_1\cup E_2 = E_1 + E_2 - (E_1\cap E_2)$$ Since $E_1,E_2$ are both measurable, it follows that $E_1\cup E_2 \;,\; E_1 \cap E_2$ are both measurable. Hence $$m(E_1 \cup E_2)=m(E_1)+m(E_2)-m(E_1 \cap E_2)$$ $$\implies m(E_1 \cup E_2) + m(E_1 \cap E_2)=m(E_1)+m(E_2)$$

Was this deduction valid or should there be more rigor? If more rigor, how would I go about doing so? I know I could also write $$m(E_1 \cup E_2) = m((E_1\cup E_2)\cap E_1)+m((E_1\cup E_2)\cap E_1^c)$$ and play from there, but things seem to get quite convoluted. Any help would be much appreciated!

Savage Henry
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  • I dont know much about sets but there might be some axioms in this which may help you in solving you problem and if those axioms are known then please let me know using a link or stating it here – Jasser Sep 16 '14 at 06:18
  • Your working is applicable when the sets involved are finite. If the sets involved have measures infinity, then you might run into the case of infinity minus infinity. – Novice Sep 16 '14 at 06:48
  • What do you mean by $E_1+E_2$? – Gerry Myerson Sep 16 '14 at 07:09
  • Your steps are not justified. I'd suggest using the following: $$E_1 \cup E_2 = (E_1 - E_2) \cup (E_2 - E_1) \cup (E_1 \cap E_2)$$ This has the advantage of the terms being disjoint. – GWilliams Sep 16 '14 at 07:28
  • Here is the link to my Math Stack Exchange post just a while ago on this very problem: https://math.stackexchange.com/questions/4158555/prob-24-chap-2-in-roydens-real-analysis-m-lefte-1-cup-e-2-rightm-left – Saaqib Mahmood Jun 02 '21 at 04:54

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When you use $+$ in $$E_1 \cup E_2 = E_1 + E_2 − (E_1 \cap E_2),$$ are you saying that $E_1 \cup E_2$ equals the disjoint union of $E_1$ and $E_2-(E_1 \cap E_2)$?

Assuming so, you may need to justify the following intermediate steps in order to be fully rigorous: $$m\left(E_1 \cup E_2\right) = m\left(E_1\right) + m\left(E_2-(E_1 \cap E_2)\right)$$ $$m\left(E_2-(E_1 \cap E_2)\right) = m\left(E_2\right) - m\left(E_1 \cap E_2\right)$$

Also, as Novice points out in the comments, you would need to separately consider infinity in this approach.

Moh
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