Linearity of integral

Let $f: {\bf R}^n \rightarrow {\bf R}$ and

$\displaystyle\int_Q f(n) dV = \displaystyle\lim_{m \rightarrow \infty} \displaystyle\sum_i f( e_i ) \Delta V$, a $(n+1)$-dimensional volume,

where $\Delta V$ an $n$-dimensional volume of an $n$-dimensional box.

Sum rule of integral. Integration preserves addition.

Let $f, g$ continuous. Then

$$\begin{array}{} \int_Q ( f(u) + g(u) ) dV &= \lim_{i \rightarrow \infty} \sum_i ( f(c_i) + g(c_i) ) \Delta V \\ &= \lim_{i \rightarrow \infty} ( \sum_i f(c_i) \Delta V + \sum_i g(c_i) \Delta V ) \\ &= \lim_{i \rightarrow \infty} \sum_i f(c_i) \Delta V + \lim_{i \rightarrow \infty} \sum_i g(c_i) \Delta V {\rm \hspace{3pt} (sum \hspace{3pt} rule \hspace{3pt} of \hspace{3pt} limits)} \\ &= \int_Q f(u) dV + \int_Q g(u) dV. \end{array}$$

Homogeneity of integral. Integration preserves scalar multiplication.

$$\begin{array}{} \int_Q k f(u) dV &= \lim_{m \rightarrow \infty} \sum_i k f(c_i) \Delta V \\ &= k \lim_{m \rightarrow \infty} \sum_i f(c_i) \Delta V \\ &= k \int_Q f(u) dV \end{array}$$

These properties suggest that the integral is a linear operator in some sense?

Fix $Q \subset {\bf R}^n$, the region of integration, then

$$\int_Q f(u) dV = H(f) \in {\bf R}, {\rm \hspace{3pt} so \hspace{3pt} that}$$

$H: C(Q ) \rightarrow {\bf R}$ is a function, $C( Q )$ is the set of all continuous functions $f: Q \rightarrow {\bf R}$, which is a vector space (of infinite dimension).

In this sense $H$ is a linear operator.