Exercise: signed measures, total variation. 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.
 A: Consider the following lemma:

If $\nu$ is a signed measure and $\lambda$, $\mu$ are positive measures such that $\nu=\lambda-\mu$, then $\lambda\geqslant \nu^+$ and $\mu\geqslant \nu^-$.

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}
A: First let's convince ourselves that the equality cannot hold: let $\nu_1$ be a non-trivial measure, i.e. $|\nu_1|(A) > 0$ for some $A$, and let $\nu_2 = -\nu_1$. Then clearly $$0 = |\nu_1 + \nu_2|(A) < 2|\nu_1|(A).$$

You make a mistake when looking at $$(\nu_1^+(A)+\nu_2^+(A))-(\nu_1^-(A)+\nu_2^-(A))$$
you conclude that $$(\nu_1^+(A)+\nu_2^+(A)) = (\nu_1 + \nu_2)^+(A)$$$$(\nu_1^-(A)+\nu_2^-(A)) = (\nu_1 + \nu_2)^-(A).$$
Indeed as long as $$(\nu_1^+(B)+\nu_2^+(B)) - (\nu_1^-(B)+\nu_2^-(B)) > 0$$ we have that $B \subset P$, where $P$ is the positive set in the HJD of $\nu_1 + \nu_2$.
This concludes the answer to your question.

If you are still interested in proving the inequality you should first convince yourself that the following holds.

Let $P_i,\ N_i$ be the HJD of $\nu_i$ and let $P,\ N$ be the HJD of $\nu_1 + \nu_2$. Then we have $$P_1 \cap P_2 \subset P \subset P_1 \cup P_2,$$ $$N_1 \cap N_2 \subset N \subset N_1 \cup N_2$$ and there are cases when the inclusions are strict.

Then 
\begin{align}
|\nu_1 + \nu_2|(A) = & (\nu_1 + \nu_2)^+(A) + (\nu_1 + \nu_2)^-(A) \\
= & (\nu_1 + \nu_2)(P\cap A) - (\nu_1 + \nu_2)(N\cap A) \\
\le & \nu_1(P_1\cap A) + \nu_2(P_2\cap A) - \nu_1(N_1\cap A) - \nu_2(N_2\cap A) \\
 = & \nu_1^+(A) + \nu_1^-(A) + \nu_2^+(A) + \nu_2^-(A) \\
 = & |\nu_1|(A) + |\nu_2|(A).
\end{align}
This proves the result. $\blacksquare$
