Timeline for Setwise convergence of induced measure converging weakly
Current License: CC BY-SA 4.0
9 events
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Mar 28, 2022 at 16:28 | comment | added | Michael Greinecker | @Gioppa Even being constant would not be enough. The sequence of functions with constant value $1/n$ converges uniformly, but the distributions do not converge setwise. | |
Mar 28, 2022 at 15:31 | comment | added | Gioppa | @MichaelGreinecker I was looking for conditions on the functions $f_n$ and $f$. Maybe smoothness of something similar. | |
Mar 28, 2022 at 15:29 | history | edited | Gioppa | CC BY-SA 4.0 |
added 21 characters in body
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Mar 28, 2022 at 15:28 | comment | added | Gioppa | I apologize for the notation, thanks for the corrections. | |
Mar 26, 2022 at 17:32 | history | edited | Yuval Peres | CC BY-SA 4.0 |
corrected notation per first comment by D. Kadelka.
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Mar 25, 2022 at 17:01 | comment | added | Michael Greinecker | Generally, $\int g~\mathrm d\mu_f=\int g\circ f~\mathrm d\mu$, so pointwise convergence implies weak convergence of the distributions by the dominated convergence theorem. So the question boils down to conditions for pointwise convergence to imply setwise convergence of distributions. Setwise convergence of distributions seems to be too strong a requirement to be satisfied in any nontrivial situation. | |
Mar 25, 2022 at 16:42 | comment | added | Dieter Kadelka | Your notation is irritating. You use $f$ for different types of functions, f.i. in $\int fd\mu_f$. | |
Mar 25, 2022 at 15:33 | history | edited | Gioppa | CC BY-SA 4.0 |
added 30 characters in body
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Mar 25, 2022 at 15:27 | history | asked | Gioppa | CC BY-SA 4.0 |