# Interchanging limit and integral of difference quotient

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 $$f$$

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

• 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. – Alex Ortiz Apr 12 at 3:22
• 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] – Jandré Snyman Apr 12 at 9:20
• 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/… – Alex Ortiz Apr 12 at 18:31