= Solution
Put $A=\int_0^1 f$. Then
$$
Rf(x)=Ax-\int_0^x f(s)\,ds.
$$
The theorem on <differentiation of an indefinite Lebesgue integral> makes this an <absolutely continuous function> on $[0,1]$, with
$$
\boxed{(Rf)'=A-f\text{ almost everywhere},\qquad Rf(0)=Rf(1)=0.}
$$
Both $A-f$ and $Rf$ belong to the <L1 space>: for example, $|Rf(x)|\leq|A|+\|f\|_{L^1}\leq2\|f\|_{L^1}$. The ordinary <derivative> of an <absolutely continuous function> is its <weak derivative>, as <integration by parts> against a <test function> shows. Therefore $\boxed{Rf\in W^{1,1}(0,1)}$, the indicated <Sobolev space>.
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