Lipschitz equivalent definition.

Question

Is Lipschitz continuous equivalent to this definition, f is Lipschitz continuous on [a,b] if for each $\epsilon > 0$ there exists $\delta > 0$ such that for every finite collection of intervals $\{(a_i,b_i)\}_{i=1}^n$ of open intervals in $(a,b)$, if $\sum_{i=1}^n [b_i-a_i] < \delta$ then $\sum_{i=1}^n |f(b_i)-f(a_i)|< \epsilon$.

Answer 1

Since $f$ is absolutely continuous we can write $f(x)=\int\limits_{0}^{x}g(t)dt$ for some integrable function $g.$ Let $x$ be a Lebesgue point of $g.$ Taking $a_{j}=x,b_{j}=x+\frac{\delta }{2n}$ for $% 1\leq j\leq n$ (the same interval is repeated $n$ times) we get $\sum\limits_{j=1}^{n}\left\vert f(x+\frac{\delta }{% 2n})-f(x)\right\vert <\epsilon .$ Hence $n\left\vert \int\limits_{x}^{x+\delta /(2n)}g(t)dt\right\vert <\epsilon .$ This holds for all $n$ and since $x$ is a Lebesgue point of $g$ we get $\left\vert g(x)\right\vert \leq \frac{2\epsilon }{\delta }.$ We have proved that $g$ is an $L^{\infty }$ function. Hence $f$ is Lipschitz.