Timeline for Compactness with respect to topology induced by total-variation distance
Current License: CC BY-SA 4.0
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| Sep 23 at 11:32 | vote | accept | Andrew Luo | ||
| Sep 22 at 11:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 11:39 | comment | added | Iosif Pinelis | @AndrewLuo : See the two additions to the answer. The second one of them shows that no real-valued measurable function $\varphi$ will guarantee that your set of measures be compact wrt to the TV metric. | |
| Sep 22 at 11:35 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 11:29 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 11:20 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 11:14 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 11:08 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 10:56 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 10:49 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 10:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 10:23 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Sep 22 at 10:06 | comment | added | Michael Greinecker | @AndrewLuo The same counterexample works if you take $\phi$ to be the bounded continuous function given by $\phi(x)=\max\{0,1-d(x,[0,1])\}$ | |
| Sep 22 at 9:03 | comment | added | Andrew Luo | Thank you for the counterexample. I am thinking that maybe I could change the class of test functions, and take $\varphi$ to be continuous bounded measurable. | |
| Sep 22 at 1:49 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |