Timeline for Is a semicontinuous real function Borel measurable?
Current License: CC BY-SA 3.0
4 events
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| Mar 10, 2012 at 5:32 | comment | added | Rami | I did not explained why it is enough to show that $(x|g(x) < c)$ is Borel. I now understand that is probably the point that you where interested in. But it is explained in the answer of GH so there is no point to repeat it | |
| Mar 10, 2012 at 5:23 | comment | added | Rami | In-fact the division of $U$ into 2 sets is unnecessary and the image of $U$ is just open. This dose not prove continuity yet since it is not enough to check continuity on sets like this. | |
| Mar 10, 2012 at 5:03 | history | edited | Rami | CC BY-SA 3.0 |
added 6 characters in body
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| Mar 10, 2012 at 4:56 | history | answered | Rami | CC BY-SA 3.0 |