## Integral over a subset

### Set

 context $\mathbb K = \overline{\mathbb R}\lor \mathbb C$ context $\langle X,\Sigma,\mu_X\rangle$ … measure space
 definiendum $\int_S: \mathcal P(X)\to(X\to \mathbb K)\to \mathbb K$
 range $f: X\to \mathbb K$
 definiendum $\int_S\ f\ \mathrm d\mu_X:=\int_X\ f\cdot \chi_S\ \mathrm d\mu_X$

### Discussion

If $X=\mathbb R$, $a,b\in \mathbb R$, $a<b$ and the measure $\mu_X$ is such that single points have zero measure $\mu_X(\{a\})=\mu_X(\{b\})=0$ (like the standard Lebesgue measure), then we write

$\int_a^b\ f\ \mathrm d\mu_X\equiv\int_{[a,b]}\ f\ \mathrm d\mu_X$

The zero measure of $a,b$ guaranties that we replace integrals over $[a,b)$, $(a,b]$ and $(a,b)$ by this one.

If $c,d\in \mathbb R$ are numbers with $c<d$, then if we write integral symbol $\int_d^c$ (notice the switched positions of $c$ and $d$ w.r.t. their ordering) we mean the negative of the integral over $[c,d]$

$\int_d^c\ f\ \mathrm d\mu_X\equiv -\int_c^d\ f\ \mathrm d\mu_X$