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Oct 14, 2023 at 19:28 comment added Julian Newman Replacing $\mu\Big(\Big[\frac{j}{k},\frac{j+1}{k}\Big)\Big)^{\!\epsilon}$ with $\, \mathbf{1}_{[\epsilon,\infty)}\!\!\left(\mu\Big(\Big[\frac{j}{k},\frac{j+1}{k}\Big)\Big)\right)$ should make it work.
Oct 14, 2023 at 18:35 comment added Christian Remling If $d\mu = \chi_{(0,1)}\, dx$, then the sum equals $k^{1-\epsilon}$, so the first limit is already $=\infty$.
Oct 14, 2023 at 18:06 review Low quality posts
Oct 14, 2023 at 19:38
Oct 14, 2023 at 17:43 history answered Anthony Quas CC BY-SA 4.0