I have this exercise:
Let $\nu_1$ and $\nu_2$ be finite signed measures on
$(\Omega,\mathcal{A})$. Prove that:$|\nu_1+\nu_2|\le|\nu_1|+|\nu_2|$;
, that is $|\nu_1+\nu_2|(A)\le|\nu_1|(A)+|\nu_2|(A)$ for each $A \in
\mathcal{A}$.
My problem is that I get an equality and not an inequality, so that leads me to believe I have done something wrong, because that is a stronger resulat than they ask for. Can you see where I did wrong, or should there be an equality?
Proof:
We have that $(\nu_1+\nu_2)(A)=\nu_1(A)+\nu_2(A)$. Then use the Jordan decomposition theorem on each part:
$=\nu_1^+(A)-\nu_1^-(A)+\nu_2^+(A)-\nu_2^-(A)=(\nu_1^+(A)+\nu_2^+(A))-(\nu_1^-(A)+\nu_2^-(A))$. This gives us the Jordan-decomposition of $\nu$ since it's Jordan-decomposition is unique.
Hence we have that the total variation is:
$|\nu_1+\nu_2|(A)=(\nu_1+\nu_2)^+(A)+(\nu_1+\nu_2)^-(A)=_\text{by what showed above}$
$(\nu_1^+(A)+\nu_2^+(A))+(\nu_1^-(A)+\nu_2^-(A))=\nu_1^+(A)+\nu_1^-(A)+\nu_2^+(A)+\nu_2^-(A)$
$=|\nu_1|(A)+|\nu_2|(A)$
Do you see an error?, I get equality, but it should be inequality. I doubt they would have written the exercise like that if in fact it should be equality.
Best Answer
Consider the following lemma:
To see this, let $(P,N)$ be a Hahn decomposition for $\nu$, then for $E\subset P$, $$\nu(E) = \nu^+(E) = \lambda(E) - \mu(E)\implies \lambda(E) = \nu^+(E)+\mu(E)\geqslant \nu^+(E), $$ and similarly if $E\subset N$, $$\nu(E) = -\nu^-(E) = \lambda(E) - \mu(E)\implies \mu(E) = \nu^-(E) + \lambda(E)\geqslant \nu^-(E), $$ since $\lambda,\mu\geqslant0$.
Now, we have \begin{align} \nu_1+\nu_2 &= (\nu_1^+ - \nu_1^-) + (\nu_2^+-\nu_2^-)\\ &= (\nu_1^++\nu_2^+)-(\nu_1^-+\nu_2^-), \end{align} so $(\nu_1^++\nu_2^+)\geqslant (\nu_1+\nu_2)^+$ and $(\nu_1^-+\nu_2^-)\geqslant (\nu_1+\nu_2)^-$. It follows that \begin{align} |\nu_1+\nu_2| &= (\nu_1+\nu_2)^+ + (\nu_1+\nu_2)^- \\ &\leqslant (\nu_1^++\nu_2^+) + (\nu_1^-+\nu_2^-)\\ &= (\nu_1^++\nu_1^-) + (\nu_2^++\nu_2^-)\\ &= |\nu_1| + |\nu_2|. \end{align}