Timeline for How to show that x-y is Lebesgue-Lebesgue measurable
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 24, 2013 at 1:44 | comment | added | mStudent | see Stein and Shakarchi, `Real Analysis' Chapter 2. | |
| Mar 31, 2010 at 17:59 | comment | added | Gerald Edgar | By coincidence (?) we just came to this exact exercise in: G. Folland, Real Analysis. This is Exercise 5 on page 245. I won't post more until after the class discusses it... | |
| Mar 26, 2010 at 22:37 | vote | accept | Nicolò | ||
| Mar 26, 2010 at 22:56 | |||||
| Mar 26, 2010 at 13:31 | answer | added | Robin Chapman | timeline score: 19 | |
| Mar 26, 2010 at 12:57 | answer | added | Colin McQuillan | timeline score: 6 | |
| Mar 26, 2010 at 12:18 | comment | added | Matthew Daws | Yep, I completely agree! Sorry! And I agree that Robin should make this an answer. | |
| Mar 26, 2010 at 12:00 | comment | added | François G. Dorais | @Robin: Please make your comments into an answer. | |
| Mar 26, 2010 at 11:36 | comment | added | Nicolò | Robin is right, every continuous function is Lebesgue-Borel measurable, but it is not said to be Lebesgue-Lebesgue measurable. My problem arise from showing that if f is measurable, then is f(x-y). Unlucky if two function $f, g$ are measurable (i.e. Lebesgue-Borel measurable), their composition $f\circ g$ not needs to be. It is if the $g$ is Lebesgue-Lebesgue measurable | |
| Mar 26, 2010 at 11:28 | comment | added | Robin Chapman | Another example: the embedding $\mathbb{R}\to\mathbb{R}^2$ taking $x$ to $(x,0)$ isn't Lebesgue-Lebesgue measurable in Nicolo's sense. | |
| Mar 26, 2010 at 10:50 | comment | added | Robin Chapman | Continuity does not imply measurability in Nicolo's strong sense. There are continuous bijections mapping sets of positive measure to sets of zero measure (e.g. Cantor sets). Each subset of a zero-measure Cantor set has Lebesgue measure zero but its inverse image need not be measurable. Decomposing the subtraction map as a composite of $(x,y)\mapsto(x-y,x)$ and $(x,y)\mapsto x$ does it cleanly enough for me. The usual definition of a Lebesgue measurable function requires the inverse image of a Borel set to be Lebesgue integrable: this is weaker than Nicolo's condition. | |
| Mar 26, 2010 at 10:40 | comment | added | Matthew Daws | Erm, isn't d continuous, and hence automatically measurable? | |
| Mar 26, 2010 at 10:29 | history | asked | Nicolò | CC BY-SA 2.5 |