This is question is spin-off of my prior question: Trouble with change of variables when constructing a smooth bump function localized on a given open ball, but this one is more suited to understanding how to properly use the change of variables theorem. Wikipedia states that the change of variables theorem is as follows:
Let $f$ be a Lebesgue measurable function and $U$ be a Lebesgue measurable subset of $\mathbb{R}^n$ and $\phi:U\to\mathbb{R}^n$ be an injective function. Suppose that for every $x\in U$ there exists $\phi'(x)$ in $\mathbb{R}^{n,n}$ such that $\phi(y) = \phi(x) + \phi'(x)(y-x) + o(||y - x||)$ as $y\to x$ (with little o notation). Then $\varphi(U)$ is Lebesgue measurable and for any real-valued function $f$ defined on $\phi(U)$: $$\int_{\phi(U)}f(v)dv = \int_U f(\phi(u))|\mathrm{det}(\phi'(u))|du$$
So let's get back to Trouble with change of variables when constructing a smooth bump function localized on a given open ball:
Now our initial integral is given by
$$g(z) = \chi_{B(x, 2r)}\ast f(z) = \int_{B(x, 2r)}f(z - y)dy$$
with
$$f(y) = r^n\varphi(\frac{y - x}{r})$$
for $L^1$-normalized
$$\varphi(x) = \begin{cases}c\exp\left(\frac{-1}{1 - |x|^2}\right) &: |x| < 1\\\ 0 &: |x|\geq 1\end{cases}$$
i.e. $\int_{\mathbb{R}^n}\varphi(x)dx = 1$. We would like to show that $g(z) = 1$ for all $z\in B(x,r)$.
We can note that $f$ is a symmetric function, so
$$g(z) = \chi_{B(x, 2r)}\ast f(z) = \int_{B(x, 2r)}f(y - z)dy$$
Problem(s) start:
1.) So at this point, going along the change of variables theorem, do we take $\phi(y) = y - z$ with the interpretation that $v := y - z$ and $u = y$? If so, then
$$B(x, 2r) = \phi(U), \phi^{-1}(w) = w + z$$
so
$$U = \phi^{-1}(B(x, 2r)) = B(x + z, 2r)$$
and
$$g(z) = \int_{B(x + z, 2r)}f(\phi(y))dy$$
2.) If step 1.) was done by the book, then the following is not clear to me: We are still looking to show that $g(z) = 1$. We know that the integral of the smooth bump function over the unit ball centered at the origin is 1, so we will have to massage the domain of integration to resemble something of the like.
So now,
$$g(z) = \int_{B(x + z, 2r)}f(\phi(y))dy = \int_{B(x + z, 2r)}r^{-n}\varphi\left(\frac{y - x}{r}\right)dy$$
Again, shifting by $x$ by reassigning
$$\phi(y) := y - x, U := \phi^{-1}(B(x + z, 2r)) = B(z, 2r)$$
we get
$$g(z) = \int_{B(z, 2r)}r^{-n}\varphi\left(\frac{y}{r}\right)dy$$
and then, by scaling by $r$,
$$\phi(y) := y/r, d\phi(y) = r^{-n}dy, U := \phi^{-1}(B(z, 2r)) = B(r\cdot z, 2r)$$
we get
$$g(z) = \int_{B(r\cdot z, 2r)}\varphi(y)dy$$
And at this point, I'm a bit lost on how to "massage" last domain of integration to be anything like the unit ball centered at the origin.