# Defining Lebesgue measure on a subspace of $\mathbb{R}^n$

Let $\bar{w}_1,.., \bar{w}_k$ be linearly independent vectors in $\mathbb{R}^n$. Let $W$ be the subspace spanned by these $\bar{w}_i$'s. I know how the Lebesgue measure is defined on $\mathbb{R}^n$. How does one define Lebesgue measure on $W$ and assign volumes to measurable subsets of $W$?

My attempt for clarification by an example: In $\mathbb{R}^3$, take $\bar{e}_1 = [1,0,0]$ and $\bar{e}_2 = [0,1,0]$. Let $V = Span \{\bar{e}_1,\bar{e}_2 \}$. The Lebesgue measure $m$ on $\mathbb{R}^3$ computes the volume of measurable sets in $\mathbb{R}^3$ and it gives $m(V) = 0$. But we can also define Lebesgue measure on $V$, which computes the area of measurable subsets of $V$. My question is about how one can generalize this example from $\mathbb{R}^3$ to $\mathbb{R}^n$ and to subspace $W$.

PS This came up when I was reading something related to geometry of numbers where they computed volumes of certain parallelepipeds in the lattice of a subspace of $\mathbb{R}^n$. And I was wondering how it worked. Thank you!

PPS I would also appreciate any reference.

• If $k < n$ then $W$ has zero ($n$-dimensional) Lebesgue measure. If $k = n$ then $W = \mathbb{R}^n$. Please clarify what you want to ask about. – user225318 Apr 18 '15 at 20:49
• The Lebesgue measure of a strict subspace of $\mathbb{R}^n$ is $0$. W is the contable union of box of measure 0 – Tryss Apr 18 '15 at 20:50
• I fixed it. I hope my question is more clear now. Thank you! – Oga Apr 18 '15 at 21:10
• Some keywords: Hausdorff measure or integration on (sub)manifolds. – PhoemueX Apr 18 '15 at 21:22