Equivalent definition of absolutely continuous

Solution 1:

Assume the second condition holds for some $\delta$ and $\varepsilon$ and choose some disjoint intervals $(c_k,d_k)$ with $\sum\limits_kd_k-c_k\lt\delta$. Then $\sum\limits_+d_k-c_k\lt\delta$ where $\sum\limits_+$ indicates that the sum is restricted to the indices $k$ such that $f(d_k)\gt f(c_k)$. In particular $\sum\limits_+ |f(d_k)-f(c_k)|=\sum\limits_+ f(d_k)-f(c_k)\lt\varepsilon$.

Likewise $\sum\limits_- |f(d_k)-f(c_k)|=-\sum\limits_- f(d_k)-f(c_k)=\left|\sum\limits_- f(d_k)-f(c_k)\right|\lt\varepsilon$, where $\sum\limits_-$ indicates that the sum is restricted to the indices $k$ such that $f(d_k)\leqslant f(c_k)$.

Summing these two contributions, one sees that the first condition holds for $\delta$ and $2\varepsilon$.

The other implication is direct.