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!
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.