For any subspace $V$ of $\Bbb R^n$, we have a special measure $\lambda_V$ which can be described in various ways: Haar measure on $V$, or the measure induced by the metric $V$ inherits from $\Bbb R^n$, or "$k$-dimensional Lebesgue measure on $V$" when $\dim V=k$. I want to be better able to calculate with this measure.

For example, I might have linearly independent vectors $v_1,v_2\in \Bbb R^n$ and $V={}$span$\langle v_1,v_2 \rangle$. Suppose I have an integral of the form $$ \int_{\Bbb R} \int_{\Bbb R} F(tv_1+uv_2) g_1(t) g_2(u) \,du \,dt, $$ where $F\colon \Bbb R^n\to\Bbb R$ and $g_1,g_2\colon \Bbb R\to\Bbb R$. Can I say that this integral is equal to $$ \int_V F(v) h_1(v) h_2(v) \,d\lambda_V, $$ for some functions $h_1,h_2\colon V\to\Bbb R$? Perhaps each $h_j$ is $g_j$ composed with some canonical projection from $V$ onto the vectors it's made from. (And perhaps there needs to be some global constant corresponding to the determinant of something involving $\{v_1,v_2\}$.)

I emphasize that I'm looking for a way to rigorously establish the connection between those two integrals; my formal ability with these Lebesgue measures on subspaces lags behind my intuition. Bonus points for an explanation that includes a reference to where these Lebesgue measures on subspaces are concretely defined and discussed.

(This related question discusses a special case of the measure $\lambda_V$, but no answer was provided.)

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    $\begingroup$ A real problem here is that Haar measure is only defined up to a constant. You need to fix a $k$-volume for the second of these integrals to be well defined. $\endgroup$ – user24142 May 15 '15 at 15:10

One difficulty is that if $n\geq 3$ then the first integral makes sense as a nonzero value, but the second is an integral over a 2-d plane which has measure 0 in more than 2 dimensions. I see that is why you mention "$k$-dimensional Lebesgue measure" if $\dim(V) = k$.

In that case, perhaps it is useful to use the standard change-of-variable formula. Let $h_1$ and $h_2$ be an orthonormal basis for $V$. So any vector $v \in V$ can be written $v = ah_1 + bh_2$. Define $M$ as the (invertible) matrix that maps $(t,u)$ to the $(a,b)$ coefficients of $v = tv_1 + uv_2$. So:

\begin{align} [a;b] &= M[t;u]\\ dadb &= |det(M)|dtdu \end{align} Thus: \begin{align} &\int_{\mathbb{R}}\int_{\mathbb{R}}F(tv_1 +uv_2)g_1(t)g_2(u)dudt \\ &=\int_{\mathbb{R}}\int_{\mathbb{R}}F\left([h_1 \: h_2] M[t;u]\right)g_1(t)g_2(u)dudt \\ &=\int_\mathbb{R}\int_{\mathbb{R}} F([h_1 \: h_2][a;b])g_1([1 \: 0]M^{-1}[a;b])g_2([0 \: 1]M^{-1}[a;b])\frac{1}{|det(M)|}dadb\\ &=\int_V F(v)h_1(v)h_2(v)d\lambda_v \end{align} where the second equality is the standard change-of-variables formula, and the last equality defines: \begin{align} h_1(v) &= \frac{g_1([1 \: 0]M^{-1}[a(v);b(v)])}{\sqrt{|det(M)|}}\\ h_2(v) &= \frac{g_2([0 \: 1]M^{-1}[a(v);b(v)])}{\sqrt{|det(M)|}} \end{align} where $a(v)$ and $b(v)$ are the $(a,b)$ coefficients associated with the orthonormal representation $v = ah_1 + bh_2$.

Here I am defining the measure of any (measurable) subset $C \subseteq V$ as: $$ \lambda(C) = \int\int_{\{(a,b): ah_1 + bh_2 \in C\}} dadb $$


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