For \(n\ge1\) and \(0\le j\le2^n-1\), let \[ I_{n,j}=[j2^{-n},(j+1)2^{-n}]. \] For \(f\in L^1([0,1])\), define \[ E_nf(x)=\sum_{j=0}^{2^n-1}\left(2^n\int_{I_{n,j}}f(t)\,dt\right)\chi_{I_{n,j}}(x). \] Prove that \(E_nf(x)\to f(x)\) for almost every \(x\in[0,1]\).
Solution
Except at dyadic endpoints, for each \(n\) there is a unique interval \(I_n(x)\) of length \(2^{-n}\) containing \(x\), and \[ E_nf(x)=\frac1{|I_n(x)|}\int_{I_n(x)}f(t)\,dt. \] The intervals \(I_n(x)\) shrink to \(x\) and satisfy \[ I_n(x)\subset[x-2^{-n},x+2^{-n}]. \] If \(x\) is a Lebesgue point of \(f\), then \[ \begin{aligned} |E_nf(x)-f(x)| &\le \frac1{|I_n(x)|}\int_{I_n(x)}|f(t)-f(x)|\,dt\\ &\le 2\cdot\frac1{2^{1-n}}\int_{x-2^{-n}}^{x+2^{-n}}|f(t)-f(x)|\,dt. \end{aligned} \] The last expression tends to \(0\) by the Lebesgue differentiation theorem. Since almost every point of \([0,1]\) is a Lebesgue point of \(f\), and the dyadic endpoints form a countable set, we obtain \[ E_nf(x)\longrightarrow f(x) \] for almost every \(x\in[0,1]\).