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Let $(X,\mathcal{F})$ be a measurable space. Let $\mu : \mathcal{F} \rightarrow \mathbb{R}$ be a real measure (i.e. $\mu (\phi) = 0$ and $\mu$ is $\sigma$-additive). Let $|\mu| (E) = \sup \left\{\sum_{h=0}^\infty |\mu(E_h)|, E_h\; {\rm are~pairwise~disjoint},\; \bigcup_{h=0}^\infty E_h = E\right\}.$ If $\mu^+ = \frac{|\mu| + \mu}{2}$ and $\mu^-=\frac{|\mu| - \mu}{2}$ then is it true that :

$\int_X u\, d\mu^+ + \int_X u \, d\mu^- = \int_X u \, d|\mu| \qquad \text{for all } |\mu|\text{-measurable }u : X \rightarrow [0,\infty]?$

I am doubtful if this is a standard result, if so can you point to any references?

The notation used here is from the book : "Functions of Bounded Variation and Free Discontinuity Problems" by Luigi Ambrosio et. al.

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    Thanks. I will have to refresh my memory about the mct.2011-05-17

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I just sum up the comments.

Approximation $u$ pointwise by a non-decreasing sequence $\{u_k\}$ of simple functions. The wanted equality is a consequence of the definition of the Lebesgue integral when $u$ is the characteristic function of a measurable set. So by linearity, we deduce that for each $k$, $\int_Xu_k\mathrm d\mu^++\int_Xu_k\mathrm d\mu^-=\int_Xu_k\mathrm d|\mu|.$ Now use monotone convergence in order to take the limit $k\to +\infty$.