I have shown that $$ T_h(x)= \begin{cases} x+h & x\in[0,1-h]\\ x+h-1& x\in(1-h,1] \end{cases} $$ for some fixed $h\in(0,1)$ is measurable and measure preserving on $([0,1],\mathcal{B}_{[0,1]},\lambda^1|_{[0,1]})$. Using this I want to show that for differentiable $f$ we have $$ \int_0^1f'(x) dx = f(1)-f(0). $$
Now using the measure preserving nature of $T_h$ I can show that $$ \lim_{h \rightarrow 0}\int_0^1 \frac{f(x+h)-f(x)}{h} dx = f(1)-f(0) $$
Now I am wanting to use the dominated convergence theorem to swap the order of the limit and integral. Now I am not sure if there exists an integrable function $w \in \mathcal{L}^1$ so that $$ \left|\frac{f(x+h)-f(x)}{h}\right| \leq w(x) $$ for all $h$. If there exists such a $w$ then I can use the dominated convergence theorem to swap the order of the limit and the integral and then I am done.
EDIT: As per the first comment on this question, I don't think this approach will work. How can I show, by using the measure preserving nature of $T_h$ that for differentiable $f$
$$ \int_0^1f'(x) dx = f(1)-f(0). $$