Here is Prob. 26, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
Let $\left\{ E_k \right\}_{k=1}^\infty$ be a countable disjoint collection of measurable sets. Prove that for any set $A$, $$ m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right) = \sum_{k=1}^\infty m^* \left( A \cap E_k \right). $$
My Attempt:
If there is a positive integer $k_0$ for which $m^* \left( A \cap E_{k_0} \right) = \infty$, then by the monotonicity of outer measure we have $$ \begin{align} m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right) &= m^* \left( \bigcup_{k=1}^\infty \left( A \cap E_k \right) \right) \\ &\geq m^* \left( A \cap E_{k_0} \right) \\ &= \infty, \end{align} $$ and hence $$ m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right) = \infty. $$ On the other hand, as $m^* \left( A \cap E_k \right) \in [0, \infty]$ for every positive integer $k$, therefore for every positive integer $n > k_0$, we have $$ \sum_{k=1}^n m^* \left( A \cap E_k \right) \geq m^* \left( A \cap E_{k_0} \right), $$ and thus we also have $$ \begin{align} \sum_{k=1}^\infty m^*\left( A \cap E_k \right) &= \lim_{n \to \infty} \sum_{k=1}^n m^*\left( A \cap E_k \right) \\ &\geq m^* \left( A \cap E_{k_0} \right) \\ &= \infty, \end{align} $$ and hence $$ \sum_{k=1}^\infty m^*\left( A \cap E_k \right) = \infty. $$ Thus our desired identity holds.
So we assume that, for every positive integer $k$, the relation $m^* \left( A \cap E_k \right) < \infty$ holds.
By the countable subadditivity of outer measure (i.e. Proposition 3 in Royden), we have $$ \begin{align} m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right) &= m^* \left( \bigcup_{k=1}^\infty \left( A \cap E_k \right) \right) \\ &\leq \sum_{k=1}^\infty \left( A \cap E_k \right). \tag{1} \end{align} $$
Let $n$ be any positive integer. Since the sets in the countable colection $\left\{ E_k \right\}_{k = 1}^\infty$ are (pairwise) disjoint measurable sets, so too are the sets in the finite collection $\left\{ E_k \right\}_{k = 1}^n$, and hence by Proposition 6 in Royden and the monotonicity of outer measure, we have $$ \begin{align} m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right) &= m^* \left( \bigcup_{k=1}^\infty \left( A \cap E_k \right) \right) \\ &\geq m^* \left( \bigcup_{k=1}^n \left( A \cap E_k \right) \right) \\ &= \sum_{k=1}^n m^* \left( A \cap E_k \right), \end{align} $$ that is, $$ \sum_{k=1}^n m^* \left( A \cap E_k \right) \leq m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right). \tag{2} $$
Since the right-hand-side of (2) is independent of $n$, therefore (2) holds for each positive integer $n$, and thus we can conclude that $$ \begin{align} \sum_{k=1}^\infty m^* \left( A \cap E_k \right) &= \lim_{n \to \infty} \sum_{k=1}^n m^* \left( A \cap E_k \right) \\ &\leq m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right). \tag{3} \end{align} $$
Finally, (1) and (3) together yield our desired identity.
Is this proof correct and clear enough? Or, are there any shortcomings in any part thereof?
PS:
Since (2) holds for every positive integer $n$ and since the sequence $\left( \sum_{k=1}^n m^* \left( A \cap E_k \right) \right)_{n \in \mathbb{N}}$ is a monotonically increasing sequence (of non-negative real numbers), therefore we have $$ \begin{align} \sum_{k=1}^\infty m^* \left( A \cap E_k \right) &= \lim_{n \to \infty} \sum_{k=1}^n m^* \left( A \cap E_k \right) \\ &= \sup \left\{ \, \sum_{k=1}^n m^* \left( A \cap E_k \right) \mid n \in \mathbb{N} \, \right\} \\ &\leq m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right), \end{align} $$ because by (2) the (extended) real number $m^* \left( A \cap \bigcup_{k=1}^\infty E_k \right)$ is an upper bound for the set $\left\{ \, \sum_{k=1}^n m^* \left( A \cap E_k \right) \mid n \in \mathbb{N} \, \right\}$.