Timeline for $f_\epsilon=\inf\{f(y):|y-x|<\epsilon\}$ is measurable Borel [closed]
Current License: CC BY-SA 4.0
10 events
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May 18, 2021 at 21:15 | history | closed |
Gro-Tsen abx Christian Remling Andreas Blass leo monsaingeon |
Not suitable for this site | |
May 15, 2021 at 19:13 | vote | accept | Zaragosa | ||
May 15, 2021 at 14:12 | answer | added | Gerald Edgar | timeline score: 1 | |
May 15, 2021 at 7:12 | history | edited | Zaragosa |
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May 15, 2021 at 7:02 | comment | added | Zaragosa | @NateEldredge for the number 2. In the number 1 I found an error, actually I need to prove only this $f_\epsilon(z)\neq a$. | |
May 15, 2021 at 6:59 | history | edited | Zaragosa | CC BY-SA 4.0 |
added 16 characters in body
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May 15, 2021 at 0:45 | comment | added | Nate Eldredge | #1 looks fine to me. The boundedness is only needed if you want $f_\epsilon$ to end up being real-valued, and even then you would only need $f$ bounded below. If you allow $f_\epsilon$ to take the value $-\infty$ then it works for unbounded functions too. | |
May 15, 2021 at 0:15 | review | Close votes | |||
May 18, 2021 at 21:15 | |||||
May 15, 2021 at 0:13 | comment | added | Nate Eldredge | For 2, note that for each $x$, $f_\epsilon(x)$ increases as $\epsilon \downarrow 0$, so you can just take the sup over $f_{1/n}(x)$, $n=1,2,3,\dots$. Then it's a sup of a countable set of functions and you are in business. | |
May 14, 2021 at 23:42 | history | asked | Zaragosa | CC BY-SA 4.0 |