The problem is Excercise 5. in page 88 of Folland's "real analysis: modern techniques and their applications", 2nd edition, as the image below shows.
As the hint indicates, we should use Excercise 4. From Excercise 4, if signed measure , then . Supposing , , we have . Then it follows . But at the same time , , so also . I really have no idea how to employ Excercise 4. to prove . Can you help me? Thanks in advance!
Thank you very much, Opalg!
After I analysised the solution, I found that the omitting condition is only used to ensure that is still a valid signed measure, so will not become which does not make sense. Based on this observation, I find it possible to extend the excercise to infinite sum of signed measures. Suppose is a sequence of signed measures on measurable space . If it happens that is still a signed measure on (it means the series is always convergent in the extended real number system on every measurable set E, and assumes at most one of . But these two points still does not ensure countable additivity(right?)), I proved that (Note that since 's are all (positive) measures, is always a valid measure). But this time Exercise 4 will not work: for each j, so we have , which, however, does not necessarily equals since the latter two series may not converge (it does in finite case). So we can not express as a difference of two measures to employ Ex.5. But I can still prove the assertion as follows: First we can prove that for any signed measure on , for any in . This is because . Assuming is a Hahn decomposition of , for , we have
in which and . So the above sum (because each term is nonegative, this is valid in ) (finite additivity of measure ) . Then we obtained the desired assertion by the arbitrariness of E.
Could you please tell me if I am right?