Originally Posted by

**Plato** Not really.

If $\displaystyle \mathbb{I} = \left\{ {\left[ {s,r} \right] \subset \left[ {a,b} \right]} \right\}$ is a collection of non-overlapping subintervals such that $\displaystyle \sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\ell \left( {\left[ {s,r} \right]} \right)} < \delta $ then

$\displaystyle \sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\left| {f(r) - f(s)} \right|} \leqslant \sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {M\ell \left( {\left[ {s,r} \right]} \right)} = M\sum\limits_{\left[ {s,r} \right] \in \mathbb{I}} {\ell \left( {\left[ {s,r} \right]} \right)} < M\delta = \varepsilon $