How come does the author deduce that from:$$\frac{\epsilon}{2^i}\sum\limits_{k \in T}\Delta x_k < \frac{\epsilon}{4^i}$$We would get:$$\sum\limits_{k \in T}\Delta x_k < \frac{\epsilon}{2^i}$$It is clear that from the first inequality we would get:$$\sum\limits_{k \in T}\Delta x_k < \frac{1}{2^i}$$However, if we suppose his deduction to be true, then we would need:$$\forall \ \epsilon > 0 : \frac{\epsilon}{2^i} < \frac{1}{2^i} \text{ or } \frac{\epsilon}{2^i} > \frac{1}{2^i}$$Since both of them do not hold, then I would assume the author has simply made an error here. It is easily fixed once we replace:$$ \omega_f(x) \geq \frac{\epsilon}{2^i} $$with:$$ \omega_f(x) \geq \frac{1}{2^i} $$Please let me know if I am missing something.