Close to diagonal implies derivative is close to 1? A function $f:[a,b]\to\mathbb R$ is absolutely continuous if it has a derivative $f'$ almost everywhere with respect to Lebesgue measure, $f'$ is Lebesgue integrable, and $f(x)=f(a)+\int_a^x f'(t)dt$ for all $x\in[a,b]$.
For an absolutely continuous homeomorphism $f:[0,1]\to[0,1]$ does being close to the diagonal imply $f'$ is, on average, close to $1$?  Precisely, if there is some $B\in[0,1]$ so that $|f(x)-x|\leq B$ does it follow that $\int_0^1|f'(x)-1|dx\leq g(B)$ where $g(B)\to 0$ as $B\to 0$?
This feels true to me, but I haven't been able to prove or disprove it.  Any help is appreciated.

No, this isn't true.  Fix a large integer $N$ and consider a function $f$ which satisfies $f(k/N)=k/N$ for all integers $k$ between $0$ and $N$, and between $k/N$ and $(k+1)/N$ is piecewise linear, with one piece of slope $1/2$ and one piece of slope $2$.  You can see that $f$ is absolutely continuous and $|f(x)-x|<1/N$ for all $x$, but $|f'(x)-1|\geq 1/2$ everywhere $f'$ is defined.
