Timeline for Local Uniform Convergence
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2016 at 19:18 | vote | accept | Hans | ||
| Apr 2, 2015 at 6:25 | comment | added | Fedor Petrov | @Hans yes, $[0,a]$ should be $[0,4a]$, "may not be disjoint" is "can not be disjoint", i.e. it is impossible that they are disjoint. Continuity allows to work with pontwise inequalities (instead of almost everywhere inequalities), it is itself essential simplification. | |
| Apr 2, 2015 at 6:17 | history | edited | Fedor Petrov | CC BY-SA 3.0 |
typo fixed
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| Apr 1, 2015 at 22:55 | comment | added | Hans | This proof as you say uses the $E_n$ and equivalently $f$ being measurable with $\mu$ being the measure function. I would rather prefer the proof to use the continuity of $f$ as requested in the question. Also, should it not be $[0,a] = \bigcup_n E_n$ rather then $[0,4a]$? And thus, all subsequent coefficients of $a$ be divided by $4$? By "B and b+B may not be disjoint" do you actually mean "... are not disjoint"? | |
| Apr 1, 2015 at 17:25 | history | answered | Fedor Petrov | CC BY-SA 3.0 |