Timeline for The relationship between measurability and weak measurability
Current License: CC BY-SA 4.0
10 events
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| Jul 7, 2022 at 0:15 | comment | added | Guomin Liu | @GeraldEdgar It seems to me that the analysis in the finite dimension ($X$ is finite dimension) of proving $L^1$ space is complete still applies by making use the completeness of $X$. But I am not quite sure. Moreover, maybe there will be measurability issue for the norm $||f||_X$ at this case, we can assume that the weak measurability test is only from a separable subspace of the dual space. | |
| Jul 6, 2022 at 16:37 | comment | added | Guomin Liu | @GeraldEdgar So if we denote by $L_s$, $L$ and $L_w$ the spaces of integrable (i.e., with finite $L^1$ norm) strongly measurable, measurable and weakly measurable functions, respectively. It is well known that $L_s$ is a Banach space. $L$ is even not a linear space since the sum of two measurable functions may not be measurable. Is $L_w$ also a Banach space? First it is easy to check that $L_w$ is a normed linear space, but is it complete? | |
| Jul 3, 2022 at 15:13 | comment | added | Guomin Liu | @GeraldEdgar Okay, I see. So the strong and weak measurability should be the right notions in the general non-separable spaces. Thanks for your comment. | |
| Jul 3, 2022 at 13:29 | comment | added | Gerald Edgar | It seems what you call "measurable" is a useless notion (when $X$ is not separable). For example, we cannot prove the sum of two measurable functions is measurable. With both "strongly measurable" and "weakly measurable", we can prove that. See mathoverflow.net/a/313792/454 and mathoverflow.net/q/294728/454 | |
| Jul 3, 2022 at 12:10 | comment | added | Guomin Liu | @MaximilianJanisch I think your proof is right. This proof is also what I think before. But I haven’t find any reference that writing this property explicitly, which makes me confused. Thanks a lot for your kind reply! | |
| Jul 3, 2022 at 9:16 | comment | added | Maximilian Janisch | If $f:\Omega\to X$ is measurable (in the preimage sense), and $g\in X^*$, then for any Borel-measurable $B\subset\mathbb R$, we have that $g^{-1}(B)$ is $\text{Borel}(X)$-measurable, so $(g\circ f)^{-1}(B)=f^{-1}(g^{-1}(B))$ is $\Omega$-measurable. So we have weak measurability. Am I missing something? | |
| Jul 3, 2022 at 7:46 | history | edited | Guomin Liu | CC BY-SA 4.0 |
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| Jul 3, 2022 at 7:14 | history | edited | Guomin Liu | CC BY-SA 4.0 |
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| Jul 3, 2022 at 6:40 | history | edited | Guomin Liu | CC BY-SA 4.0 |
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| Jul 3, 2022 at 6:26 | history | asked | Guomin Liu | CC BY-SA 4.0 |