Siegel's lattice integral formula says that for $f \in L^1 (\mathbb R^n-[{0}])$, $$\int_{\mathbb R^n} f(x) dx = \int_{L_n} f^{*}(\lambda)$$ where on the right we integrate over the covolume lattices space $L_n$ and $f^*(L) = \sum_{h\in L-{[0]}} f(h)$.
The proof goes by showing for $f \in C_c(\mathbb R^n-[{0}])$ that we have a positive linear functional using the right side(it takes some work to show $f^* \in L^1(L_n)$ for $f \in C_c(\mathbb R^n-[{0}])$- basically one shows that for indicator of balls the right side is bounded). Riez Radon then gives us that there is a measure on $f \in L^1 (\mathbb R^n-[{0}])$ that is $SL_n(\mathbb {R})$ invariant, and thus it must be a multiple of the borel measure.
I have two questions:
What is the easiest way to show the constant is $1$?
What is the easiest way to now extend this to all of $L^1$ from $C_c$?
First for $2$, I have a very clumsy approach. First you use monotone convergence to obtain this is true for indicators of open sets. Then given a measure $0$ set $E$ you approximate it from above to show $f^*(1_E)$ to see it's measureable on the right (in the completion) and the formula holds. Finally for a general measureable set you approximate it from above by open sets up to measure $0$. Is there a standard better approach?
Next for 1, I know the trick should be to take $f = 1_B / Vol(B)$ for larger and larger balls. Then pointwise $f^* \to 1$ by gausses theorem, but we cannot apply DCT because we don't have anything to dominate the right side. I could go through a similiar analysis as before, bounding the error in gauss theorem based on a represenative in the Siegel set and showing the error is small, but shouldn't there be an easier way?