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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 $f$

$$ \int_0^1f'(x) dx = f(1)-f(0). $$

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  • $\begingroup$ Functions like $f(x) = x^2\sin(1/x)$ for $x > 0$, and $f(x) = 0$ for $x = 0$, make me doubt that we can apply the dominated convergence theorem here. The derivative of this function $f$ I give is probably not in $L^1([0,1])$, though I haven't checked it: the derivative has a term with a factor of $1/x$ in it. $\endgroup$ – Alex Ortiz Apr 12 at 3:22
  • $\begingroup$ I'm Not sure that you're right about the function you define, it has derivative 2xsin(1/x)-cos(1/x) which is bounded and therefore integrable on [0,1] $\endgroup$ – Jandré Snyman Apr 12 at 9:20
  • $\begingroup$ Sorry, I dropped a power of $2$: the function I meant is $f(x) = x^2\sin(1/x^2)$. I was reading from here: mathcounterexamples.net/… $\endgroup$ – Alex Ortiz Apr 12 at 18:31

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