Skip to main content
9 events
when toggle format what by license comment
Jun 12 at 9:48 vote accept Vassilis Papanicolaou
Jun 12 at 9:48 vote accept Vassilis Papanicolaou
Jun 12 at 9:48
Jun 9 at 2:25 comment added Iosif Pinelis @GregMartin : In this case, $f$ may not be Riemann integrable. So, unfortunately, I still don't see how the definition for $x<0$ can be avoided.
Jun 7 at 17:29 comment added Greg Martin For Riemann integrals the definition is identical (it just happens that $\Delta x < 0$ in the Riemann sums). Lebesgue integrals are typically taken over sets rather than oriented sets.
Jun 7 at 12:52 comment added Iosif Pinelis @GregMartin : Thank you for your comment. How is it well defined, if not as in this answer?
Jun 7 at 7:40 comment added Greg Martin Nice argument; note that there's no need for the second definition of $F(x)$ since $\int_0^x f(t)\,dt$ is still well defined when $x<0$.
Jun 6 at 17:00 history edited Iosif Pinelis CC BY-SA 4.0
added 3 characters in body
Jun 6 at 15:05 history edited Iosif Pinelis CC BY-SA 4.0
added 22 characters in body
Jun 6 at 14:53 history answered Iosif Pinelis CC BY-SA 4.0