I'm trying to understand the proof of existence for the maximal coupling. That is:
For any two probability measures $\mathbb{P},\mathbb{P}'$ on a measurable space $(E,\mathcal{E})$ there exists a maximal coupling $\hat{\mathbb{P}}$, such that equality is obtained in the coupling inequality $$ \|\mathbb{P}-\mathbb{P}'\|_{tv} \leq 2\hat{\mathbb{P}}(\hat{X}=\hat{X}'). $$
Where $\| \cdot \|_{tv}$ is the total variation norm.
Now before I go on I first need to define the following:
put $ \lambda = \mathbb{P}+\mathbb{P}'$, then $g = \frac{d\mathbb{P}}{d\lambda}$ and $g' = \frac{d\mathbb{P'}}{d\lambda}$.
Now in the proof it is stated that by using the Jordan-Hahn decomposition of signed measures into a difference of non-negative measures we find that:
$$ \|\mathbb{P}-\mathbb{P}'\|_{tv} = \int_E |g-g'|d\lambda $$
I have looked at the Jordan-Hahn decomposition, and I grasp the concept, but I do not see how this equality follows. If anyone could help me I would really appreciate it!