# Thread: A question on a signed measure inequality

1. ## 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!

2. Originally Posted by zzzhhh
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 $\nu_2^\pm$.
Originally Posted by zzzhhh
... 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 $-\infty$.

3. 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?

4. 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.

5. 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?

6. Originally Posted by zzzhhh
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.