# Equivalent definition of absolutely continuous

A function $f$ is absolutely continuous on $[a,b]$ is defined by: for each $\varepsilon>0$, there is a $\delta>0$, for each finite disjoint open interval $\{(c_k,d_k)\}_{k=1}^n$ contained in $[a,b]$, we have $$\text{if}\,\, \sum_{k=1}^n (d_k-c_k)<\delta, \,\,\text{then}\,\, \sum_{k=1}^n\left|f(d_k)-f(c_k)\right|<\varepsilon.$$

However, in the book I'm reading, it is said that there is a equivalent definition, say $$\text{if}\,\, \sum_{k=1}^n (d_k-c_k)<\delta,\,\, \text{then}\,\, \left|\sum_{k=1}^n f(d_k)-f(c_k)\right|<\varepsilon.$$

However, I can not prove it. How?

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Hint: the absolute value of the sum is less or equal to the sum of the absolute values. – Nana May 30 '12 at 14:54
For the other direction, consider partitions where $f(d_k)\geq f(c_k)$, $\forall k$, and $f(d_k) < f(c_k)$, $\forall k$. – copper.hat May 30 '12 at 15:13

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 $2\varepsilon$.