The law induced by a norm $\rho$ on $\mathbb{R}^n$ with respect to Lebesgue's measure. The found the following problem in C. George's Exercices et problèmes d'intégration, Problem 3.65:

Suppose $\rho$ is a norm on  $\mathbb{R}^n$. Let $V=\lambda_d(\rho\leq1)$, where $\lambda_n$ denotes Lebesgue measure on $(\mathbb{R}^n,\mathscr{B}(\mathbb{R}^n))$.  Show that
$$
\int_{\mathbb{R}^n}f(\rho(x))\,dx=nV\int^\infty_0 t^{n-1}f(t)\,dt\tag{1}\label{one}
$$
for any integrable $f\in L_1([0,\infty),\mathscr{B}([0,\infty)),x^{n-1}\,dx)$, and more generally, for any measurable positive measurable function on $[0,\infty)$.
Hint:  If $E\subset[0,\infty)$ has Lebesgue measure $0$, then $\lambda_n(\{x\in\mathbb{R}^n: \rho(x)\in E\})=0$. To see this,  show that for any set $E\subset[0,b]$ of measure zero, there is a sequence $\{f_m: m\in\mathbb{B}\}$ of step functions with common compact support such that
$$0\leq f_m\leq f_{m+1},\qquad\sup_m\int^\infty_0f_m <\infty,\qquad\text{and}\quad f_m\xrightarrow{m\rightarrow\infty}\infty\quad\text{on}\quad E$$


I have a direct solution to this problem based on monotone class arguments: For any closed bounded interval $I$, the space $\mathcal{V}_I$ of all bounded Borel measurable functions supported in $I$ for which $\eqref{one}$ holds is a vector space closed under taking limits of uniformly bounded monotone sequences. Since $\rho$ is homogeneous of order 1 and $\lambda_n(aV)=a^n\lambda_n(V)$ for any measurable set $V\subset\mathbb{R}^n$, $\mathcal{V}_I$ contains the multiplicative class $\mathcal{M}_I$ of step functions supported in $I$. Hence $\mathcal{V}_I$ contains all bounded functions measurable with respect to $\sigma(\mathcal{M})=\mathscr{B}(I))$. $\Box$

I however, would like to see a solution relying on the hint.
What I am struggling is to  extend the identity to integrable functions $f$ (w.r.t $\mu(dx)=x^{n-1}\,dx$), in particular showing that if $f_n\rightarrow f$  almost surely in $\lambda_n$ ($\lambda_n$-a.s.), then $f_n\circ\rho\rightarrow f\circ \rho$ $\lambda_n$ a.s.
I think the key is to show that $\lambda_d(\rho\in E)=0$ for any set $E\subset[0,\infty)$ with $\lambda_1(E)=0$.

I will appreciate if  someone can provide a proof of the fact in the hint.
 A: Here is a prove of the statement of the hint.
Keeping the notation of the problem,  $\lambda_n$ notes the $n$-th dimensional Lebesgue measure, and  $V$ denotes the volume of the unit $\rho$-ball $\{x\in\mathbb{R}^n: \rho(x)\leq1\}$. Suppose $E\subset[0,\infty)$ is a set with $\lambda_1(E)=0$. Since $E=\bigcup^\infty_{n=1}E\cap[n,n+1)$, it is enough to show that $\lambda_n(\{\rho\in E\})=0$ for bounded sets $E$ of $\lambda_1$-measure $0$.
On any interval $[a,b]\subset[0,\infty)$, the function $g(x)=x^n$ is absolutely continuous. Then, for any $\varepsilon>0$, there is $\delta>0$ such that for any subintervals $\{I_j=[\alpha_j,\beta_j]:j\in\mathbb{N}\}$ in $[a,b]$,
$\sum_j|g(\beta_j)-g(\alpha_j)|<\varepsilon/V$ whenever $\sum_j(\beta_j-\alpha_j) <\delta$.
Since $\rho$ is a norm, it follows that $\rho$ is Borel measurable, and by homogeneity
$$\lambda_n(\{x\in\mathbb{R}^n:\rho(x)\leq r\})=r^n V$$
for any $r\geq0$.
If $r>0$, $\{\rho=r\}\subset\{(1-\frac{1}{m}r<\rho\leq r\}$. Hence
$$\lambda_n\big(\{\rho=r\}\big)\leq\lambda_n\big(\{(1-\frac{1}{m}r<\rho\leq r\}\big)=(r^n-(1-\tfrac1m)^nr^n)V\xrightarrow{m\rightarrow\infty}0$$
As a consequence, for any $0\leq a\leq b<\infty$
$$\lambda_n\big(\{a<\rho\leq b\}\big)=\lambda_n\big(\{a\leq \rho\leq b\}\big)=(b^n-a^n)V$$
To conclude the proof, suppose $E\subset [a,b]$ has $\lambda_1$-measure $0$. Then, there is a sequence of  intervals $I_j=[\alpha_j,\beta_j]$ in $[a,b]$ such that $E\subset\bigcup_jI_j$ and $\sum_j\lambda_1(I_j)<\delta$. This implies that
$$\lambda_n\big(\{\rho\in E\}\big)\leq\lambda_n\left(\bigcup_j\{\rho\in I_j\}\right)\leq\sum_j\lambda_n\big(\{\rho\in I_j\})=V\sum_j(g(\beta_j)-g(\alpha_j))<\varepsilon$$
As $\varepsilon$ can be chosen to be arbitrarily small, the conclusion is that $\lambda_n(\{\rho\in E\})=0$.
