Lebesgue measure on finite dimensional inner product space

Question

Let $V$ be $d$ finite dimensional inner product space and $<.,.>_{V}$ be inner product on it. Let $v_1, \dots, v_{d}$ denotes its orthonormal basis. Let $\lambda$ denote Lebesgue measure on $\mathbb{R}^{d}$, then define Lebesgue measure $\mu$ on $V$ using following rule,

Let $E$ be any subset of $V$ and then define its measure $\mu(E) = \lambda(\vec{E})$ where $\vec{E} = \{ c| c \text{ is coordinate of some e in } E \}.$

Now we are trying to integrate a function $f : V \rightarrow \mathbb{R}$ using this measure $\mu$ on $V.$ My question is that

Can we convert $f$ in coordinates and then integrate in $\mathbb{R}^{d}$ under Lebesgue measure ?

Ex. consider $f(v) = <v,v>_{V}$, now assume that we can write inner product in coordinates $<v,v>_{V} = \vec{v} G \vec{v}$ where $\vec{v}$ are coordinates of $v$ and $G$ is basis matrix. Does following hold?

$$\int_{V} <v,v>_{V}\, d \mu \overset{?}{=} \int_{\mathbb{R}^d} \vec{v}G^{T} \vec{v}\, d \lambda.$$

Answer 1

The general setting is as follows.

Let $(X, \mathscr{X})$ and $(Y, \mathscr{Y})$ be two measure spaces and $\varphi:X \to Y$ be a $(\mathscr{X}, \mathscr{Y})$-measurable function. Given a measure $\mu$ on $(X, \mathscr{X}),$ one can induce a measure on $Y$ by $F \mapsto \mu(\varphi^{-1}(F)).$ Such a measure is called the "image measure of $\mu$ by $\varphi$" and denoted $\varphi(\mu).$ By definition then $$ \int \mathbf{1}_F\ d\varphi(\mu) = \int \mathbf{1}_F \circ \varphi\ d\mu. $$ The usual way shows then that $$ \int f\ d\varphi(\mu) = \int f \circ \varphi\ d\mu, $$ for every function $f:Y \to \mathbf{R}$ for which the previous integrals make sense (the conclusion is that both exists in the extended sense and are equal or both do no exist).

What you are considering is $X = \mathbf{R}^d$ with the Borel sets, and $Y$ a $d$-dimensional vector space with the Borel sets (note that the Borel sets of a finite dimensional vector space are uniquely defined). And $\varphi = [\cdot]_\beta^{-1}$ for a basis $\beta$ given in advance. The above construction shows $$ \int f \circ [\cdot]_\beta^{-1}\ d\mu = \int f\ d[\mu]_\beta^{-1}. $$ This formula holds for all Borel measurable functions $f:Y \to \mathbf{R}.$ So, for example, if you have $f:Y \to \mathbf{R}$ being an inner product, and you can write it as $f(v) = [v]_\beta^\intercal G [v]_\beta,$ then $f \circ [\cdot]_\beta^{-1}:x \mapsto x^\intercal G x,$ which answers affirmatively your question.