# About the sum of two integrals and Jordan decomposition

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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I'm not sure what you're asking (I'm a bit puzzled about your emphasis "$u$ is $|\mu|$-measurable"), but it looks a bit like if you're asking whether $|\mu| = \mu^{+} + \mu^{-}$ holds, but that is obviously true by definition. Note also that $\mu^{+}, \mu^{-} \ll |\mu|$ are positive measures and $\mu^{+} \perp \mu^{-}$. –  t.b. May 17 '11 at 6:05
@The: I wanted to know if the sum of integrals on the left equals the integral on the right with some restrictions on $u$. I could prove that for simple functions but not for general $u : X \rightarrow [0,\infty]$. Hope this is more clear. –  jpv May 17 '11 at 6:13
But yes, as $u \geq 0$, it can be approximated monotonically by a sequence of simple functions, so you can simply apply the monotone convergence theorem on both sides of the identity, as you're only dealing with positive measures and functions. Or am I missing something? –  t.b. May 17 '11 at 6:18
Thanks. I will have to refresh my memory about the mct. –  jpv May 17 '11 at 6:24

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 $k\to +\infty$.