The idea is to bound the two expressions tightly in terms of each other: $ \sum_{k=-\infty}^{+\infty} 2^k m(F_k) \leqslant \sum_{k=-\infty}^{+\infty} 2^k m(E_k) \leqslant 2 \cdot\sum_{k=-\infty}^{+\infty} 2^k m(F_k) . $ Once this is established, it is obvious that either summation converges if and only if the other does,
The left half of the inequality is obvious since $F_k \subseteq E_k$, so we prove only the right half.
$ \begin{align*} \sum_{k=-\infty}^{+\infty} 2^k m(E_k) &\leqslant \sum_{k=-\infty}^{+\infty} 2^k \left( \sum_{j=k}^{\infty} m(F_j) \right) \\ &= \sum_{j= -\infty}^{\infty} m(F_j) \left( \sum_{k=-\infty}^{j} 2^k \right) \\ &= \sum_{j= -\infty}^{\infty} 2^j m(F_j) \left( \sum_{k=-\infty}^{0} 2^k \right) \\ &= \sum_{j= -\infty}^{\infty} 2^j m(F_j) \cdot 2. \end{align*} $ Here the interchange of summations is justified because all terms are nonnegative.