Here is Prob. 24, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
Show that if $E_1$ and $E_2$ are measurable, then $$ m \left( E_1 \cup E_2 \right) + m \left( E_1 \cap E_2 \right) = m \left( E_1 \right) + m \left( E_2 \right). $$
My Attempt:
As the sets $E_1$ and $E_2$ are measurable (subsets of $\mathbb{R}$), so are their union and intersection.
And, as the sets $E_1 \cap E_2$ and $E_2$ are measurable, so is the set $E_2 \setminus \left( E_1 \cap E_2 \right)$.
If $m \left( E_1 \cap E_2 \right) = m^* \left( E_1 \cap E_2 \right) < \infty$, then since $E_1 \cap E_2 \subset E_2$, therefore using the excision property we obtain $$ m \left( E_2 \setminus \left( E_1 \cap E_2 \right) \right) = m \left( E_2 \right) - m \left( E_1 \cap E_2 \right), $$ and since we can write $$ E_1 \cup E_2 = E_1 \cup \left( E_2 \setminus \left( E_1 \cap E_2 \right) \right), $$ and since the sets $E_1$ and $E_2 \setminus \left( E_1 \cap E_2 \right)$ are disjoint, therefore by Proposition 6 in Royden we also have $$ m \left( E_1 \cup E_2 \right) = m \left( E_1 \right) + m \left( E_2 \setminus \left( E_1 \cap E_2 \right) \right) = m \left( E_1 \right) + m \left( E_2 \right) - m \left( E_1 \cap E_2 \right), $$ which implies $$ m \left( E_1 \cup E_2 \right) + m \left( E_1 \cap E_2 \right) = m \left( E_1 \right) + m \left( E_2 \right), $$ as required.
On the other hand, if $m \left( E_1 \cap E_2 \right) = \infty$, then for each $i = 1, 2$, as $$ E_1 \cap E_2 \subset E_i \subset E_1 \cup E_2, $$ so from the monotonicity of the (outer) measure it follows that $m \left( E_i \right) = \infty$ and also $m \left( E_1 \cup E_2 \right) = \infty$, and thus the desired identity holds trivially.
Is my proof correct and clear enough in each and every bit of detail? Or, are there any mistakes in it or issues of lack of clarity for that matter?
PS:
Here is Proposition 6, Chap. 2, in Royden:
Let $A$ be any set and $\left\{ E_k \right\}_{k=1}^n$ a finite disjoint collection of measurable sets. Then $$ m^* \left( A \bigcap \left[ \bigcup_{k=1}^n E_k \right] \right) = \sum_{k=1}^n m^* \left( A \cap E_k \right). $$ In particular, $$ m^* \left( \bigcup_{k=1}^n E_k \right) = \sum_{k=1}^n m^* \left( E_k \right). $$