# The Unapologetic Mathematician

## Integrals are Additive Over Regions

Before I state today’s proposition, I need to define what I mean by saying that two sets in $\mathbb{R}^n$ are “nonoverlapping”. Intuitively, we might think that this means they have no intersection, but that’s not quite it. We’ll allow some intersection, but only at boundary points. Since the regions we’re interested in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no difference.

Let $\left\{S_i\right\}_{i=1}^n$ be a collection of bounded regions in $\mathbb{R}^n$, so that any two of these regions are nonoverlapping. We define their union

$\displaystyle S=\bigcup\limits_{i=1}^nS_i$

and let $f:S\rightarrow\mathbb{R}$ be a bounded function defined on this union. Then $f$ is integrable on $S$ if and only if it’s integrable on each $S_i$, and we find

$\displaystyle\int\limits_Sf\,dx=\sum\limits_{i=1}^n\int\limits_{S_i}f\,dx$

Indeed, if $R$ is some $n$-dimensional interval containing $S$, then it will also contain each of the smaller regions $S_i$, and we can define

$\displaystyle\int\limits_Sf\,dx=\int\limits_Rf(x)\chi_S(x)\,dx$

and

$\displaystyle\int\limits_{S_i}f\,dx=\int\limits_Rf(x)\chi_{S_i}(x)\,dx$

We can use Lebesgue’s condition to establish our assertion about integrability. On the one hand, the discontinuities of $f$ in $S_i$ must be contained within those of $S$, so if the latter set has measure zero then so must the former. On the other hand, the discontinuities of $f$ consist of those within each of the $S_i$, and maybe some along the boundaries. Since the boundaries have measure zero, and we assume that the discontinuities in each $S_i$ are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of $f$ in $S$ have measure zero as well.

The inclusion-exclusion principle tells us that we can rewrite the characteristic function of $S$:

$\displaystyle\begin{aligned}\chi_S&=\chi_{S_1\cup\dots\cup S_n}\\&=\sum\limits_{1\leq i\leq n}\chi_{S_i}-\sum\limits_{1\leq i<j\leq n}\chi_{S_i\cap S_j}+\sum\limits_{1\leq i<j<k\leq n}\chi_{S_i\cap S_j\cap S_k}-\dots+(-1)^n\chi_{S_1\cap\dots\cap S_n}\end{aligned}$

We can put this into our integral and use the fact that integrals are additive with respect to finite sums in the integrand

$\displaystyle\begin{aligned}\int\limits_Sf\,dx&=\int\limits_Rf(x)\chi_S(x)\,dx\\&=\sum\limits_{1\leq i\leq n}\int\limits_Rf(x)\chi_{S_i}(x)\,dx-\sum\limits_{1\leq i<j\leq n}\int\limits_Rf(x)\chi_{S_i\cap S_j}(x)\,dx+\dots+(-1)^n\int\limits_Rf(x)\chi_{S_1\cap\dots\cap S_n}(x)\,dx\\&=\sum\limits_{1\leq i\leq n}\int\limits_{S_i}f\,dx-\sum\limits_{1\leq i<j\leq n}\int\limits_{S_i\cap S_j}f\,dx+\dots+(-1)^n\int\limits_{S_1\cap\dots\cap S_n}f\,dx\end{aligned}$

But we assumed that all the $S_i$ are nonoverlapping, so any intersection of two or more of them must lie only along their boundaries. And since these boundaries all have Jordan content zero the integrals over them must come out to zero. We’re left with only the sum over each subregion, as we wanted.

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Posted by John Armstrong | Analysis, Calculus

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