Activity for Michael
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #296664 | Initial revision | — | 4 days ago |
| Answer | — |
A: Defining a explicit function, without axiom of choice, that is not Lebesgue integrable on any interval? To get a ZF-definable function that is not Lebesgue integrable on any interval, we need a function that is measurable but has \intI |f| = \infty for every interval I. Example: Define f(x)= \begin{cases} \infty, & x\in A,\\ 0, & x\notin A, \end{cases} where A is a measurable set of infinite mea... (more) |
— | 4 days ago |