I want to answer your question with Lebesgue integrals, as you ask, but the fact is that the machinery for the measure does not take this into account at all (the measure of a subset of $\mathbb R$, using the standard measure or the counting measure, is always positive). I remember having a similar issue with signed area when learning about line integrals (absolute value signs always rub me the wrong way). The answer is that it makes the addition formula hold, even in the Lebesgue sense, in the following way:

Define $\bar\int_S\,f=-\int_Sf$. The idea is that $\bar\int$ "undoes" integrals, so if $B\subset A$, then $\int_A f+\bar\int_B\,f=\int_{A-B} f$, analogous to $\int_A f+\int_B\,f=\int_{A\uplus B} f$ if $A\cap B=\emptyset$, and $\int_A f+\bar\int_A\,f=0$. But this way, we can erase more than the whole integral: $B\subset A$ implies
$$\int_A f+\bar\int_B\,f=\bar\int_{B-A} f,$$
where now $\bar\int_{B-A} f$ is supposed to represent negative integration over the set. It's nothing but algebraic trickery, but it's supposed to make you think in terms of being able to integrate backwards over things. Thus, we can define $[a,b]=\overline{[b,a]}$, where I'm abusing my own notation here to indicate that this set is supposed to be integrated negatively, and $\int_{[a,b]} f=\bar\int_{[b,a]} f\Rightarrow\int_a^b=-\int_b^a$. It's a lot of notation for very little gain, but hopefully I've convinced you that this can be made to make some sense. It's certainly useful to be able to say $\int_0^x dt=x$ instead of $\int_0^x dt=|x|$!

**Edit:** For sums, the counting measure assigns $1$ to every point in $\mathbb Z$, but the same rule applies. For $A=[a,b)\cap\mathbb Z$, we get $\int_Af=\sum_{i=a}^{b-1}f$, and for the disjoint union rule, we have, if $B=[b,c)\cap\mathbb Z$ is disjoint with $A$,
$$\int_Af+\int_Bf=\int_{A\uplus B}f\Rightarrow\sum_{i=a}^{b-1}f+\sum_{i=b}^{c-1}f=\sum_{i=a}^{c-1}f.$$

(Note how the $-1$'s appear. This is an artifact of the definition of a sum as $\sum_a^b=\sum_{[a,b]\cap\mathbb Z}$, using the closed, rather than the half-open interval, which behaves nicer w.r.t. addition. Every programmer who writes `for (int i=0;i<N;i++)`

knows this.) Rearranging this identity, we get, for a negative sum $\overline{[a,b)\cap\mathbb Z}=(b,a]\cap\mathbb Z=[b-1,a-1)\cap\mathbb Z$,

$$\sum_{i=b}^{c-1}f-\sum_{i=a}^{c-1}f=-\sum_{i=a}^{b-1}f:=\overline\sum_Af:=\sum_{i=b}^{a-1}f.$$

I've chosen the limits on the last definition to be consistent with our addition rule, because I can now rearrange to get

$$\sum_{i=b}^{c-1}f=\sum_{i=b}^{a-1}f+\sum_{i=a}^{c-1}f,$$

which looks identical to the first addition rule, except now $a<b$, so one is a backwards sum. Thus:

$$\sum_{i=b}^{a-1}f=-\sum_{i=a}^{b-1}f\Rightarrow\sum_{i=b}^af=-\sum_{i=a+1}^{b-1}f.$$