Proving that a sequence of functions is a sequence of mollifiers. Brezis’s book gives an example of a sequence of mollifiers. Considering the following smooth function:
$$
p(x)=\left\{\begin{array}{rc}
e^{\frac{1}{\vert x\vert^2-1}},&\mbox{if}\quad \vert x\vert< 1,\\
0, &\mbox{if}\quad \vert x\vert>1.
\end{array}\right.
$$
Then he says: "we obtain a sequence of mollifiers by letting $p_n(x)=Cn^N p(nx)$ with $C=\frac{1}{\int p}$. 
-> My question is as follows: 
How to prove that $\int p_n=1$? Being more punctual: I am having difficulties in changing variables. I thank you in advance for your help.
 A: Of course you would agree with me that $$\lambda\int_a^b f(\lambda x)dx = \int_a^b f(\lambda x)d(\lambda x)=  \int_{\lambda a}^{\lambda b}f(y)dy \tag{CoV}\label{CoV}$$
a
To calculate $\int p_n$ we can note that $p_n$ is zero if $|x|>1/n$ i.e. $x$ is outside the ball $B_{1/n}(0)$ and therefore zero if $\max(|x_1|,|x_2|,\dots,|x_n|) > 1/n$ i.e. outside the $N$-dimensional 'cube around $0$' $X$ of sidelength $2/n$. So 
$$ \int_{\mathbb R^N} p_n dx= \int_{\mathbb B_{1/n}(0)} p_n dx= \int_X p_n dx = \int_{-1/n}^{1/n}\dots\int_{-1/n}^{1/n} p_n(x_1,\dots,x_N) dx_1\dots dx_N$$
Apply definition of $p_n$
$$ \int_{\mathbb R^N} p_n dx =  Cn^N\int_{-1/n}^{1/n}\dots\int_{-1/n}^{1/n} p(nx_1,\dots,nx_N) dx_1\dots dx_N$$
Now apply \ref{CoV} $N$ times. Each time uses up one copy of $n$ on the outside. So the $n^N$ disappears exactly, leaving you with 
$$ \int_{\mathbb R^N} p_n dx = C\int_{-1}^1 \dots \int_{-1}^1 p(y_1,\dots,y_N)dy_1\dots dy_N= C\int_{B_1(0)}p(y)dy = C\int_{\mathbb R^N} p(y) dy$$
now we see $C$ was chosen specifically for this to be equal to 1.
For functions that are not compactly supported you can prove a more general multivariate change of variables theorem that avoids talking about boxes.
