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Saturday, June 15, 2013

2.12

First, define the sets: S_n^k = \Big[\frac{k-1}{2^n}, \frac{k}{2^n}\Big] From these intervals, define the sequence of functions: f_n^k = \chi_{S_n^k}(x) + \chi_{S_n^k}(-x) Where for each n to progress +1, k must sweep from 1 to 2^n + 1.

Certainly, for every n, \int_{\mathbb{R}} f_n^k = \frac{1}{2^{n-1}} Since the collection of functions \lbrace f_n^k \rbrace is countable, with a well-defined sequence, let us just enumerate them with the single index m. We have: \int_{\mathbb{R}} |f_m - 0| = \int_{\mathbb{R}} f_m \to 0 However, for any x \in \mathbb{R}, it's clear that \lim_{m\to \infty} f_m(x) does not exist. To expand this to \mathbb{R}^d, just replace the intervals with a closed balls centered at the origin subtracting open balls with the same dyadic rational-difference... (onion-layers).

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