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Hi I am confused on something. We defined the lebesgue integral of a general f in the following way: If $\int f^+ <\infty $ and $\int f^- < \infty$ then $\int f= \int f^+ - \int f^-. $

So basically if we have an integral like $\int f+g $ then we can only seperate it if both f and g are integrable?

Also if we have a certain integral of f over the real numbers, can we split that integral into the sum of integrals over sets $A_i$ whose union is the real numbers? Can we do this even if we dont know whether the function is integrable? Thanks!

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    What sense would it make to separate the integral into non-defined integrals (since the functions are not integrables or the domains not measurables)? – 2017-02-17
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    Thanks, basically I am wondering how to show a Lebesgue integral like $\int_{[0,1]} 1/x^2 $ , say, does not exist, if I can not seperate it into $[0,\epsilon]$ and $[\epsilon,1]$ and taking limits. Could somebody clarify? – 2017-02-17
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    If you want to show it does not exist, you can do so assuming it does, then split it and reach a contradiction. – 2017-02-17
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    oh right so if we get infinity then thats a contradiction? Thank you. – 2017-02-17

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Note that functions that aren't lebesgue integrable may still sum to a lebesgue integrable function: just take any non-integrable $f$ and let $g = -f$; neither are lebesgue integrable, but their sum is the zero function, which is lebesgue integrable.

Furthermore, if $f$ is a complex valued lebesgue integrable map on a measure space $(X,\mathbf{M},\mu)$, then the map

$\tilde{\mu}: \mathbf{M} \to \mathbb{C} \,;\, E \mapsto \int_E fd\mu$

will be a complex measure on $\mathbf{M}$, which is to say that it is complex valued and countably additive. Consequently, if $A_1,A_2,... \in \mathbf{M}$ are pairwise disjoint measurable sets then

$\int_{\bigcup_iA_i} f d\mu = \tilde{\mu}\left( \bigcup_i A_i \right) = \sum_i \tilde{\mu}(A_i) = \sum_i \int_{A_i} f d\mu$

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    Thanks, so if f is integrable it holds. But I was wondering if we can not seperate the integral into different integrals, then how we could show a Lebesgue integral like $\int_{[0,1]} 1/x^2 $ , say, does not exist, if I can not seperate it into $[0,\epsilon]$ and $[\epsilon,1]$ and taking limits. Could somebody clarify? – 2017-02-17
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    You could try and find a sequence $s_1 \leqslant s_2 \leqslant \cdots \leqslant |f|$ of simple measurable functions $s_n:X \to [0,\infty)$ that converges pointwise to $|f|$ but is such that $\int_X s_n d\mu \to \infty$. It would follow that $f$ is not lebesgue integrable. – 2017-02-17