Lemma 2

If $J$ is a finite subset of $I$ such that $\varphi(J) < 1/4$, then

For since $\sigma_L \leq \sigma_J$ for all subsets $L$ of $J$, it follows from (2) that $|s_L| \leq \varphi(J) \sigma_J + |p_L - 1| \leq (1 + \sigma_J)\varphi(J)$; but by virtue of Chapter VII, $\S$ 3, no. 1, Proposition 2, we have $|\sigma_J| \leq 4 \sup_{L \subset J} |s_L|$, hence $\sigma_J \leq 4 \varphi(J)(1 + \sigma_J)$, and the result follows.

Now let us show that the condition stated in Theorem 1 is sufficient. The hypothesis that the family $(|u_i|)$ is summable in $\mathbf{R}$ implies that the family $(1 + |u_i|)$ is multipliable in $\mathbf{R}_+^*$ (Chapter IV, § 7, no. 4, Theorem 4); hence, for each $\varepsilon > 0$, there is a finite subset $J_0$ of $I$ such that, for each finite subset $L$ of $I$ which does not meet $J_0$, we have $\prod_{i \in L} (1 + |u_i|) - 1 \leq \varepsilon$. But we can write $\prod_{i \in L} (1 + u_i) - 1$ in the form $\sum_M \left( \prod_{i \in M} u_i \right)$ where $M$ runs through all non-empty subsets of $L$; and since $\left| \prod_{i \in M} u_i \right| = \prod_{i \in M} |u_i|$, we have

This proves our assertion, by virtue of Cauchy’s criterion, since $\mathbf{C}^*$ is a complete group.

We still have to show that the condition of Theorem 1 is necessary. If $(1 + u_i)_{i \in I}$ is a multipliable family in $\mathbf{C}^*$, there exists a finite subset $J$ of $I$ such that, for every finite subset $H$ of $I$ which does not meet $J$, we have $\left| \prod_{i \in H} (1 + u_i) - 1 \right| \leq 1/8$. By Lemma 2, it follows that $\sum_{i \in H} |u_i| \leq 1$ for every finite subset $H$ of $I$ which does not meet $J$, and hence the family $(|u_i|)$ is summable in $\mathbf{R}$ (Chapter IV, § 7, no. 1, Theorem 1).

The proof above applies also, mutatis mutandis, to (ordered) infinite products in certain non-commutative division rings and algebras (see Exercise 6, and Chapter IX, Appendix).