Let $\nu$ be a signed measure on $(X,\mathcal{M})$. The total variation of $\nu$ is defined as
$$ |\nu|= \nu^+ + \nu^-, $$
where $\nu = \nu^+ - \nu^-$ is the Jordan decomposition of $\nu$.
For every $B \in \mathcal{M}$, one can show that if $\nu$ is $\sigma$-finite, then
$$ |\nu(B)| = \sup \left(\int_Bfd|\nu|:f \text{ is measurable, } |f| \leq 1 \right). $$
Question:
How can I show that
$$ |\nu(B)| = \sup \left(\int_Bfd\nu:f \text{ is measurable, } |f| \leq 1 \right). $$
That is, the equality still holds with $\nu$ instead of $|\nu|$ in the integral?