Show that a function satisfying a Lipschitz condition is absolutely continuous

A function f is said to satisfy a Lipschitz condition on an interval if there is a constant M such that
|f(x)-f(y)| ≤ M|x-y| for all x and y in the interval
a)Show that a function satisfying a Lipschitz condition is absolutely continuous
b)Show that f satisfies a Lipschitz condition if one of its derivates (Say D^(+)) is bounded
Thanks for your help

a) Suppose f is Lipschitz on [a,b] and let e > 0. Set d = e/M. If P = {[xi, yi]} is any partition of [a,b] with ||P|| < d, then ∑ |f(xi) - f(yi) ≤ ∑ M|xi - yi| = M||P|| < M(e/M) = e. Thus f is absolutely continuous on [a,b].

b) If f is not Lipschitz, then are sequences {x_n} and {y_n} in [a,b] such that |f(x_n) - f(y_n)| > n|x_n - y_n| for all n ≥ 1. Thus |f(x_n) - f(y_n)|/|x_n - y_n| > n for all n ≥ 1. Hence lim sup |f(x_n) - f(y_n)|/|x_n - y_n| = ∞ and lim inf |f(x_n) - f(y_n)|/|x_n - y_n| = ∞. This the one sided derivatives of f are not bounded.