Lemma $3.5$ If $E_1 \subset \mathbb{R}^{d_1}$ and $E_2 \subset \mathbb{R}^{d_2}$, then $$m_* \left( E_1 \times E_2 \right) \leq m_* \left( E_1 \right) m_* \left( E_2 \right),$$ with the understanding that if one of the sets $E_j$ has exterior measure zero, then $m_* \left( E_1 \times E_2 \right) = 0$.
The above is the statement of Lemma $3.5$ in chapter $2$ of Stein's real analysis.
The author proves that $m_* \left( E_1 \times E_2 \right) \leq \left( m_* \left( E_1 \right) + \epsilon \right) \left( m_* \left( E_2 \right) + \epsilon \right)$ for any $\epsilon > 0$.
I think this completes the proof of the Lemma $3.5$ by the arbitrariness of $\epsilon > 0$.
However, in the textbook, it is divided into two cases as follows:
If neither $E_1$ nor $E_2$ has exterior measure $0$, then from the above we find $$m_* \left( E_1 \times E_2 \right) \leq m_* \left( E_1 \right) m_* \left( E_2 \right) + O \left( \epsilon \right),$$ and since $\epsilon$ is arbitrary, we must have $m_* \left( E_1 \times E_2 \right) \leq m_* \left( E_1 \right) m_* \left( E_2 \right)$.
If for instance $m_* \left( E_1 \right) = 0$, consider for each positive integer $j$ the set $E_2^j = E_2 \cap \left\{ y \in \mathbb{R}^{d_2} : \left\lvert y \right\rvert \leq j \right\}$. Then, by the above argument, we find that $m_* ( E_1 \times E_2^j ) = 0$. since $( E_1 \times E_2^j ) \nearrow \left( E_1 \times E_2 \right)$ as $j \to \infty$, we conclude that $m_* \left( E_1 \times E_2 \right) = 0$.
But, why do we have to divide the cases?
It seems to me that it suffices to describe it as the case that $E_1$ and $E_2$ has nonzero exterior measure, in both cases.
Am I missing something?
Edit) The $m_*$ is the one which is usually called Lebesgue outer measure, and called exterior measure in the textbook.
The subtlety comes from the case when, say, $m_*(E_2) = \infty$. Using completely improper, heuristic notation, the inequality you think might be sufficient would roughly state
$$ m_* \left( E_1 \times E_2 \right) \leq \left( m_* \left( E_1 \right) + \epsilon \right) \left( m_* \left( E_2 \right) + \epsilon \right) \approx(0 + \epsilon)(\infty + \epsilon).$$
It's not clear what this is supposed to mean. Then Stein and Shakarchi show that for any of the expanding sets $E_2^j$, the RHS approximation is still arbitrarily small.