Note on Besicovitch's theorem on the possible values of upper and lower derivatives
The author proves the following result. For all natural numbers \(k\), \(n_1,\dotsc,n_k\) and bases \(B_i\) in \({\mathbb R}^{n_i}\) that are Busemann-Feller, regular and complete, the basis \(B_1\times \dotsb \times B_k\) possesses the Besicovitch property. Let us recall the definitions of the concepts in the statement. A (differentiation) basis \(B\) on \(\mathbb R^n\) is a mapping defined on \(\mathbb R^n\) such that, for each \(x\in \mathbb R^n\), \(B(x)\) is a family of measurable sets of positive finite measure containing \(x\) such that there exists a sequence \(\{R_k\}\subset B(x)\) with the diameter of \(R_k\) tending to zero as \(k\) approaches infinity. For example \(\mathbb I(\mathbb R^n)\), defined such that for each \(x\in \mathbb R^n\), the family \(\mathbb I(\mathbb R^n)(x)\) consists of all \(n\)-dimensional intervals containing \(x\), is a basis. For a basis \(B\), \(\overline{B}\) shall denote the family \(\cup_{x\in \mathbb R^n} B(x)\). The basis \(B\) is said to be Busemann-Feller if \(R\in B(x)\) for all \(R\in \overline{B}\) and \(x\in R\). Let \(I_1,\dotsc,I_n\) be intervals in \(\mathbb R\) and \(I=I_1\times\dotsb \times I_n\). A basis \(B \subset \mathbb I(\mathbb R^n)\) is said to be regular if \[ \sup_{I\in \overline{B}} \frac{\max_{1\leq i\leq n} |I_i|}{\min_{1\leq i\leq n} |I_i|}<\infty. \] We say that a Busemann-Feller basis is complete if there are numbers \(0<c_1<c_2<1\) such that, for any \(I\in \overline{B}\), there exists a partition \(\{I_1,\dotsc,I_k\}\subset \overline{B}\) of the interval \(I\) with the property \[ c_1\leq \frac{|I_i|}{|I|}\leq c_2\quad i=1,\dotsc,k. \] For \(f\in L_{{\roman{loc}}}({\mathbb R}^n)\), we set \[ \overline{D}_B\Big(\int f,x\Big)=\overline{\lim} \frac{1}{|R|}\int_R f \] and \[ \underline{D}_B\Big(\int f,x\Big)=\underline{\lim}\frac{1}{|R|}\int_R f, \] where the limes inferior and superior are taken with respect to \(R\in B(x)\) with diameter of \(R\) converging to zero. A basis is said to possess the Besicovitch property if the two sets \[ \Big\{-\infty<\underline{D}_B\Big(\int f,\cdot\Big)<f\Big\}\; \text{and} \; \Big\{f<\overline{D}_B\Big(\int f,\cdot\Big)<\infty\Big\} \] have measure zero for any \(f\in L({\mathbb R}^n)\). For simplicity of exposition, the proof is carried out for the case where each \(B_i\) is such that \(B_i(x)\) consists of all \(n\)-dimensional cubic intervals containing \(x\). The proof is a refinement of the proof of a result by Ward as presented by \textit{S. Saks} [Theory of the integral. 2., revised edit. Engl. translat. by L. C. Young. With two addit. notes by Stefan Banach. New York: G. E. Stechert (1937; Zbl 0017.30004); Theory of the integral. 2. ed. English translation by L. C. Young. With two additional notes by S. Banach. New York: G. E. Stechert (1937; JFM 63.0183.05); Reprint of the 2nd revised ed. 1937. Mineola, NY: Dover Publications (1964; Zbl 1196.28001)].
Besicovitch's theorem on the values of upper and lower derivatives
scientific article; zbMATH DE number 6158185