Let $f$ be increasing on $[0,1]$ and define $g(x)=\limsup_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}$. Show that $m(A)\leq f(1)-f(0)$ where $A=\{x| g(x)>1\}$

Question

Let $f$ be increasing on $[0,1]$ and define $g(x)=\limsup_{h \to 0} \frac{f(x+h)-f(x-h)}{2h}$. Show that $m(A)\leq f(1)-f(0)$ where $A=\{x| g(x)>1\}$

I found a proof of this that uses vitali covering, However I think I found a "simpler proof," in the sense that the results I use have been proved using Vitali covering at some point. I was wondering whether my reasoning is correct.

Since $f$ is increasing it is differentiable a.e and so $g(x)=f'(x)$, and $f'\geq0$. Thus $\int_0^1f'\leq f(1)-f(0)$. But clearly then $\int_{A}f'\leq f(1)-f(0)$ and so by monotonicity $m(A)\leq f(1)-f(0)$