Multiple integral

Definition

Let $n \ge 1$ and let $B = [a_1, b_1] \times \cdots \times [a_n, b_n]$ be a closed box in $\R^n$ with $a_i < b_i$. Its volume is $\abs{B} = (b_1 - a_1) \cdots (b_n - a_n)$. A partition $P$ of $B$ is given by a partition of each interval $[a_i, b_i]$; it cuts $B$ into finitely many subboxes $B_1, \dots, B_N$. The mesh of $P$ is the largest diameter of the $B_j$. A tagged partition also chooses a point $\xi_j \in B_j$ for each $j$. For a function $f \colon B \to \R$, the Riemann sum of a tagged partition is $\sum_{j=1}^N f(\xi_j) \abs{B_j}$.

The function $f$ is Riemann integrable on $B$ if there is a real number $I$ with the following property. For each $\varepsilon > 0$ there is $\delta > 0$ such that every tagged partition of mesh less than $\delta$ has a Riemann sum within $\varepsilon$ of $I$. Then $I$ is the multiple integral of $f$ over $B$, written $$\int_B f \, dV = \int \cdots \int_B f(x_1, \dots, x_n) \, dx_1 \cdots dx_n.$$

Let $A \subseteq \R^n$ be bounded and let $f \colon A \to \R$. Let $f_A$ be the function on a closed box $B \supseteq A$ that equals $f$ on $A$ and $0$ on $B \setminus A$. Then $f$ is integrable on $A$ if $f_A$ is integrable on $B$, and $\int_A f \, dV = \int_B f_A \, dV$. This does not depend on the choice of $B$.