link : wikipedia
Consider the following standard mollifier.
$$\eta(x) = \frac{1}{z}\begin{cases} e^{-\frac{1}{1-\|x\|^2}}& \text{ if } \|x\| < 1\\ 0& \text{ if } \|x\|\geq 1 \end{cases}$$ $$\eta_\epsilon(x) = \frac{1}{\epsilon^n} \eta\left( \frac{x}{\epsilon} \right)$$
At this time, I have a question. How to derive the next equation ?
Let $y = x / \epsilon$, then $$\int_{B_\epsilon(0)} \eta_\epsilon(x) dx = \int_{B_1(0)} \eta(y) dy = 1$$
(added) $$\int_{B_\epsilon(0)} \eta_\epsilon(x) dx = \frac 1 {\epsilon^n} \int_{B_1(0)} \eta(y) (\epsilon \, dy)$$ How to divide $\epsilon^n$?
Since $x$ and $y$ are points from $R^n$, $dx$ and $dy$ represent volume element in $R^n$, i.e, $dx=dx_1dx_2\cdots dx_n$ and since $dx=\epsilon^ndy$, the cancellation takes place.