An exercise requires to show that if $f$ is a real function of bounded variation over $[0,1]$, so is $|f|^r$, where $r \geq 1$.
It is trivial to prove that $|f| \in \operatorname{BV}[0,1]$, and, should one notice that $$\operatorname{V}_0^1|f|^p \leq p \sup |f| \operatorname{V}_0^1 |f| < \infty, \quad \text{for all } p \in \mathbb{N},$$ it is straightforward that the statment is true for integer $r \geq 1$.
(The last result is obtained with the help of the inequality $\operatorname{V}_a^b (gh) \leq \sup |g| \operatorname{V}_a^b h + \sup |h| \operatorname{V}_a^b g,$ which holds for any functions $g,h : [a,b] \to \mathbb{R}$).
We have proven the case in which $r \in \mathbb{N}$. But what if $r$ is rational or even irrational?
Use the Mean Value Theorem: $$ \bigl|\,|f(x)|^r-|f(y)|^r\,\bigr|=r\,\xi^{r-1}\bigl|\,|f(x)|-|f(y)|\,\bigr|\le r\,(\sup|f|)^{r-1}|f(x)-f(y)|. $$ In the above, $\xi$ is between $|f(x)|$ and $|f(y)|$.