The paragraph above is from: Foundations of Modern Probability (1st edition) - Kallenberg (Page 11)
I am struggling to show that the extended integral as described in the attached paragraph is independent of the choice of representation. I can see that if $f = g-h$ as written, that $g \geq f^{+}$ and $h \geq f^{-}$ necessarily, and I have also derived the following formulae for $f^{+}$ and $f^{-}$ in terms of $g$ and $h$:
$$f^{+} = \frac{max(g,h) - min(g,h) + g - h}{2}$$
and $$f^{-} = \frac{max(g,h) - min(g,h) + h - g}{2}$$
but don't see an obvious application of any of these considerations to showing independence.
Uniqueness means $f=f_1-f_2=f_3-f_4$ with $f_i$'s non-negative and $\mu f_i <\infty$ for all $i$ implies $\mu f_1- \mu f_2=\mu f_3-\mu f_4$. To prove this just observe that $f_1+f_4=f_2+f_3$ and $\mu$ is linear for non-negative measurable functions. Hence $\mu f_1+\mu f_4= \mu f_2+\mu f_3$ from which the result follows.