Symmetrically differentiable functions are differentiable almost everywhere (Q790269)

Let \(R=(-\infty,\infty)\), f:\(R\to R\) and \(x\in R\). The upper (lower) symmetric derivative of f at x is \[ \bar f^ s(x)=\lim \sup_{h\to 0}\frac{1}{2h}(f(x+h)-f(x-h))\quad(\bar f^ s(x)=\lim \inf_{h\to 0}\frac{1}{2h}(f(x+h)-f(x-h))). \] A function f:\(R\to R\) is called symmetrically differentiable at x iff \(\underline f^ s(x)=\bar f^ s(x).\) A function f:\(R\to R\) is called upper (lower) symmetrically semicontinuous at x, \(x\in R\), iff \[ \lim \sup_{h\to 0+}(f(x+h)-f(x- h))\leq 0\quad(\lim \inf_{h\to 0+}(f(x+h)-f(x-h))\geq 0). \] If \(M\subset R\), then we denote by \(| M|\) the outer Lebesgue measure of M and by D(M) the set \(\{x\in R:\lim_{h\to 0+}\frac{| M\cap(x- h,x+h)|}{2h}=1\}.\) If f:\(R\to R\) and \(x\in R\), then \(\underline Df(x)=\min(\underline D^+f(x),\underline D^-f(x)),\) where Ḏ\({}^+f(x)\) and Ḏ\({}^-f(x)\) are the lower Dini derivatives of f at x. The main result is the following Lemma: If f:\(R\to R\), \(K\in R\) and E is a measurable subset of R such that \((i)\quad E\subset D(E), (ii)\quad E\subset D(\{z\in R:\underline f^ s(z)>K\})\) and (iii) f is symmtrically semicontinuous at each point of E, then \(\underline Df(z)\geq K\) at almost every point of E. The following Theorem and two Corollaries are easy consequences of this Lemma: Theorem. If f:\(R\to R\), then f is differentiable at almost all points of the set \(D(\{x\in R:\bar f^ s(x)<\infty \quad or\quad \underline f^ s(x)>-\infty \})-D(\{x\in R:\) f is not symmetrically semicontinuous at \(x\})\). Corollary 1. A function f is differentiable almost everywhere in a measurable set E if it is symmetrically differentiable almost everywhere in E. Corollary 2. If a function is symmetrically differentiable almost everywhere in R, then it is measurable.