The previous owner of the Mathematics chat Asaf Karagila drafted some etiquette rules for the chat. They were deleted, so I am reposting them so that they might be observed.
It's nice of you to drop by, after saying hello please spend a minute reading the transcript to see if there is an active...
I'm not a moderator. I have ~50k rep on Server Fault. I am a regular in chat.
Because chat allows all 10k+ users across the network to act on chat flags, I (and many other regulars in SF chat) see some...less mature...users from other chats flagging things that aren't actually offensive.
I disa...
@MDMarra If it was something along the lines of "we know", he wasn't chewing you out. It gets really exasperating dealing with people who continually misuse flags.
in the one dimensional case of the inverse function theorem, are the conditions (continuity, monotonicity) also necessary conditions for a function to have an inverse?
@FrankScience If you mean in sense of Riemann Integration, Riemann Integrable functions, that is precisely those with countable discontinuities on the domain.
@OrangeHarvester Incidentally, your proposition of Riemann integrability is not right enough. It's exactly Lebesgue-measure-zero discontinuities on the domain of a bounded function.
@WillJagy I have not learnt Lebesgue measures yet, but does not measure zero mean continuous almost everywhere? That is the impression I got from Principles of Mathematical Analysis Rudin ( I have yet to do the last chapter on measures there.)
(I understand that there is a proposition says exactly that....but will my proof still work without involkling the theorem/theorem)? (sounds silly I know ;)
