Let $f(x)=x\sin(1/x)$ for $0<x\le 1$, and $f(0)=0$. I was told that every continuous function on $[0,1]$ can be written as $a(x)+s(x)$, where $a(x)$ is absolute continuous and $s(x)$ is singular. But how to write $f(x)$ this way?
Note: I was able to verify that $f(x)$ is indeed continuous. Further, I took the derivative and it turns out it is not Lebesgue integrable on $[0,1]$, so $f(x)$ is not abs. continuous. But I am stuck to find $a(x)$ and $s(x)$.
A continuous function of bounded variation has that property. In general, a function of bounded variation can be decomposed into the sum of an absolutely continuous function, a singular function, and a jump function.
The function $x \mapsto x \sin(1/x)$ is not of bounded variation on $[0,1].$