# Lebesgue measure of $A=\{\sum_i^n t_iv_i, 0\le t_i\le 1\}$

Given are $v_1,...,v_n\in\mathbb R^n$, and the set $A=\{\sum_{i=1}^n t_iv_i, 0\le t_i\le 1\}$. How to calculate the Lebesgue measure $\lambda^n(A)$?

Can I solve it with applying $\lambda^n((a,b])=\Pi_i^n (a_i-b_i)$ $(*)$ ?

For example,$$v_1=(1,2,3)^T,v_2=(2,2,4)^T,v_3=(4,6,1)^T$$

how to apply $(*)$ ?

• You should work out some concrete examples with $n = 1$ and 2. As another hint, consider what happens when the $v_i$ are linearly dependent or when one of the vectors is scaled by a negative factor. Mar 28 '17 at 16:14
• when $v_i$ are lin.dependent then $\lambda^n(A)=0$ because $A$ isnt a $n$-dim parallelpiped, it is a hyperplane and $\lambda(hyperplane)=0$. And if one vector is scaled with a negative factor, I can define a translation $\tau:\mathbb R^n\to\mathbb R^n:\lambda^n(A)=\lambda^n(\tau (A))$ such that the vector isnt scaling with a neg.factor anymore, because the Lebesgue measure is invariant on translations. Mar 29 '17 at 9:22

The solid $A$ is the parallelpiped spanned by the vectors $v_1,\ldots,v_n$. Its Lebesgue measure is the absolute value of the determinant of the matrix whose columns are the $v_i$: $$\lambda^n(A) = \bigg| \det \begin{bmatrix} v_1 | \cdots | v_n \end{bmatrix} \bigg|.$$