Volume of a body under permutation

Question

For a permutation $\sigma :\left[ n\right] \rightarrow \left[ n\right]$ we define $T_{\sigma} \left( x \right): \mathbb{R}^n \rightarrow \mathbb{R}^n$ to be the transformation:
$T_{\sigma} \left( x \right) = \left( x_{\sigma \left( 1\right) },\ldots ,x_{\sigma \left( n\right) }\right)$, for every $x = \left( x_{1}, \ldots , x_{n} \right) \in \mathbb{R}^n$.
Show that for any body $K \subseteq \mathbb{R}^n$ it holds that $vol \left( T_{\sigma} \left( K \right) \right) = vol \left( K \right)$.
We define $vol \left( K \right) = \int _{K}1 = \int _{C} \mathbb{1}_{K}$, where $C = \left( \prod ^{n}_{i=1}C_{i} \right) \subseteq \mathbb{R}^n$ is a box (a cartesian product of segments in $\mathbb{R}$, each $C_i \subseteq \mathbb{R}$) that contains $K$, meaning $K \subseteq C$, and $\mathbb{1}_K$ is the indicator function of $K$, $\mathbb{1}_K = \begin{cases}1 \ \ \ \ x\in K\\ 0 \ \ \ \ else\end{cases}$.
I was wondering how can I prove it with upper and lower Darboux sums instead of using Fubini.