How to show a function is absolutely continuous? Are all continuous functions also absolutely continuous functions or not? If it does, then  does its inverse hold? Kindly give an example?
 A: The Cantor function (or the Devil's staircase) provides an example of a continuous function that is not absolutely continuous.
A equivalent definition of absolute continuity is that a function $f:[a,b] \to \mathbb{C}$ is absolutely continuous if there is a Lebesgue integrable function, $g : [a,b] \to \mathbb{C}$, for which $$f(x) = f(a) + \int_a^x g(\tau) d\tau.$$
Equivalently, $g(x) = f'(x)$ almost everywhere.
In the case of Cantor's function, $f'(x)$ exists and is zero almost everywhere. The integral of the zero function is always zero, and cannot increase the way that Cantor's function does.
A: A more elementary example, one that can be verified right from the definition, is $f(x) = x\sin (1/x)$ on $[0,1]$ (wth $f(0) = 0$ of course). To see this, let
$$a_n = \frac{1}{\pi/2 + 2n\pi},\, b_n = \frac{1}{2n\pi},\, I_n = [a_n,b_n].$$
Then given $\delta > 0,$ choose a positive integer $N> 1/(2\pi \delta).$ Then for $n\ge N,$ the pairwise disjoint intervals $I_n$ all lie in $[0,\delta).$ But note
$$\sum_{n=N}^{N+k}|(\Delta f)_{I_n}| = \sum_{n=N}^{N+k} \frac{1}{\pi/2 + 2n\pi}.$$
This sum can be made as large as we want by taking $k$ large. This violates the AC condition. Thus $f$ is continuous, hence uniformly continuous, on $[0,1],$ but $f$ is not absolutely continuous there.
A: Absolute continuity implies continuity, but continuity does not always imply absolute continuity. For example, the Cantor function $c:[0,1]\to [0,1]$ is uniformly continuous but it is not absolutely continuous. This is the most common example of a continuous function that is not absolutely continuous, but I believe there are others. 
