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Thread: A question on a signed measure inequality

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    A question on a signed measure inequality

    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 \nu=\lambda-\mu, then |\nu|=\nu^++\nu^-\leq\lambda+\mu. Supposing \nu_1=\lambda_1-\mu_1, \nu_2=\lambda_2-\mu_2, we have \nu_1+\nu_2=(\lambda_1+\lambda_2)-(\mu_1+\mu_2). Then it follows |\nu_1+\nu_2|\leq\lambda_1+\lambda_2+\mu_1+\mu_2. But at the same time |\nu_1|\leq\lambda_1+\mu_1, |\nu_2|\leq\lambda_2+\mu_2, so |\nu_1|+|\nu_2| also \leq\lambda_1+\lambda_2+\mu_1+\mu_2. I really have no idea how to employ Excercise 4. to prove |\nu_1+\nu_2|\leq|\nu_1|+|\nu_2|. Can you help me? Thanks in advance!
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  2. #2
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    Quote Originally Posted by zzzhhh View Post
    The problem is Exercise 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 Exercise 4. From Exercise 4, if signed measure \nu=\lambda-\mu, then |\nu|=\nu^++\nu^-\leq\lambda+\mu. Supposing \nu_1=\lambda_1-\mu_1, \nu_2=\lambda_2-\mu_2,
    In particular, you can take \lambda_1 = \nu_1^+ and \mu_1 = \nu_1^-, and similarly for \nu_2^\pm.
    Quote Originally Posted by zzzhhh View Post
    ... we have \nu_1+\nu_2=(\lambda_1+\lambda_2)-(\mu_1+\mu_2).
    Now apply Exercise 4. If \nu_1+\nu_2=(\lambda_1+\lambda_2)-(\mu_1+\mu_2), then Exercise 4 tells you that \lambda_1+\lambda_2\geqslant (\nu_1+\nu_2)^+ and \mu_1+\mu_2\geqslant (\nu_1+\nu_2)^-. Therefore

    \begin{aligned}|\nu_1|+|\nu_2| &= (\nu_1^++\nu_1^-) + (\nu_2^++\nu_2^-) \\ &= (\lambda_1+\mu_1) + (\lambda_2+\mu_2) \\ &= (\lambda_1+\lambda_2) + (\mu_1+\mu_2) \\ & \geqslant (\nu_1+\nu_2)^+ + (\nu_1+\nu_2)^- = |\nu_1+\nu_2|.\end{aligned}

    Exercise 6. Now say where you used the condition that \nu_1, \nu_2 both omit the value +\infty or  -\infty.
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    Thank you very much, Opalg!
    After I analysised the solution, I found that the omitting condition is only used to ensure that \nu_1+\nu_2 is still a valid signed measure, so \nu_1+\nu_2=(\lambda_1+\lambda_2)-(\mu_1+\mu_2) will not become \infty-\infty which does not make sense. Based on this observation, I find it possible to extend the excercise to infinite sum of signed measures. Suppose <\nu_j> is a sequence of signed measures on measurable space (X,\mathcal M). If it happens that \nu=\sum\limits_{j = 1}^\infty \nu_j is still a signed measure on (X,\mathcal M) (it means the series \sum\limits_{j = 1}^\infty \nu_j(E) is always convergent in the extended real number system \overline{\mathbb R} on every measurable set E, and \nu assumes at most one of \pm\infty. But these two points still does not ensure countable additivity(right?)), I proved that |\sum\limits_{j = 1}^\infty \nu_j|\leq \sum\limits_{j = 1}^\infty |\nu_j| (Note that since |\nu_j|'s are all (positive) measures, \sum\limits_{j = 1}^\infty |\nu_j| is always a valid measure). But this time Exercise 4 will not work: \nu_j=\nu_j^+-\nu_j^- for each j, so we have \sum\limits_{j = 1}^\infty \nu_j=\sum\limits_{j = 1}^\infty (\nu_j^+-\nu_j^-), which, however, does not necessarily equals \sum\limits_{j = 1}^\infty \nu_j^+ - \sum\limits_{j = 1}^\infty \nu_j^- since the latter two series may not converge (it does in finite case). So we can not express \nu 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 \mu on (X,\mathcal M), |\mu(E)|\leq|\mu|(E) for any E in \mathcal M. This is because |\mu(E)|=|\mu^+(E)-\mu^-(E)|\leq|\mu^+(E)|+|\mu^-(E)|=\mu^+(E)+\mu^-(E)=|\mu|(E). Assuming (P,N) is a Hahn decomposition of \nu, for \forall E\in\mathcal M, we have
    \begin{aligned}\\ &|\nu|(E)\\ &=\nu^+(E)+\nu^-(E)\\ &=|\nu^+(E)|+|\nu^-(E)|\\ &=|\nu(E\cap P)|+|-\nu(E\cap N)|\\ &=|\nu(E\cap P)|+|\nu(E\cap N)|\end{aligned},
    in which |\nu(E\cap P)|=|\sum\limits_{j = 1}^\infty \nu_j(E\cap P)|\leq\sum\limits_{j = 1}^\infty|\nu_j(E\cap P)|\leq\sum\limits_{j = 1}^\infty |\nu_j|(E\cap P) and |\nu(E\cap N)|=|\sum\limits_{j = 1}^\infty \nu_j(E\cap N)|\leq\sum\limits_{j = 1}^\infty|\nu_j(E\cap N)|\leq\sum\limits_{j = 1}^\infty |\nu_j|(E\cap N). So the above sum \leq\sum\limits_{j = 1}^\infty |\nu_j|(E\cap P)+\sum\limits_{j = 1}^\infty |\nu_j|(E\cap N)=\sum\limits_{j = 1}^\infty(|\nu_j|(E\cap P)+|\nu_j|(E\cap N)) (because each term is nonegative, this is valid in \overline{\mathbb R}) =\sum\limits_{j = 1}^\infty |\nu_j|(E) (finite additivity of measure |\nu_j|) =(\sum\limits_{j = 1}^\infty|\nu_j|)(E). Then we obtained the desired assertion by the arbitrariness of E.
    Could you please tell me if I am right?
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    Yes, that looks good. I think I would want to prove it by using Exercise 5 plus induction to show that \Bigl|\sum\limits_{j = 1}^n\nu_j\Bigr| \leqslant \sum\limits_{j = 1}^n|\nu_j|, and then letting n\to\infty provided that the limit makes sense on both sides.
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    But to use induction, we have to show that \lim\limits_{n\to \infty}|\sum\limits_{j = 1}^n \nu_j|=|\sum\limits_{j = 1}^\infty \nu_j|(=|\lim\limits_{n\to \infty} \sum\limits_{j = 1}^n \nu_j|), which I temporarily can not see how to prove. Can you prove it?
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    Quote Originally Posted by zzzhhh View Post
    But to use induction, we have to show that \lim\limits_{n\to \infty}|\sum\limits_{j = 1}^n \nu_j|=|\sum\limits_{j = 1}^\infty \nu_j|(=|\lim\limits_{n\to \infty} \sum\limits_{j = 1}^n \nu_j|), which I temporarily can not see how to prove. Can you prove it?
    Yes, you're right. I had overlooked that problem.
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