A simple proof of the Stepanov theorem on differentiability almost everywhere (Q1293799)

Let \(\Omega\subset \mathbb{R}^n\) be an open set and \(f: \Omega\to \mathbb{R}^n\) be a function. Denote by \(f'(x)\) the gradient of \(f\), i.e., the vector of all partial derivatives of \(f\) at \(x\). The function \(f\) is said to be differentiable at \(x\in\Omega\) if \(f\) has partial derivatives at \(x\) and \[ \lim_{y\to x} {f(y)- f(x)- f'(x)\cdot(y- x)\over|y-x|}= 0. \] Denote by \(\text{lip}(f,x)\) the value \(\lim_{y\to x}{|f(y)- f(x)|\over|y-x|}\) and put \(S(f)= \{x\in\Omega: \text{lip}(f,x)<\infty\}\). A very simple and elegant proof is given for the following classical theorem due to Stepanov: Let \(f\) be an arbitrary function on an open set \(\Omega\subset \mathbb{R}^n\). Then \(f\) is differentiable almost everywhere in \(S(f)\). The proof uses Rademacher's theorem stating that any Lipschitz function of an \(n\)-dimensional variable is a.e. differentiable.