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\begin{document}
Let $\nu_1$ and $\nu_2$ be valuations on the noetherian local domain $(R,m)$, with centers $p_2\subset_{\ne} p_1$ and such that $\nu_1$ is composed with $\nu_2$ and their ranks differ by one. Let $\lambda\colon \Gamma_1\to \Gamma_2$ be the corresponding map of value groups.  Set $t_1=\nu_1 (p_1), t_2=\nu_2(p_2)$.\par
Given $\phi_2\in \Gamma_2$, denote by $\tilde\phi_2\in \Gamma_1$ the minimum of $\{\nu_1 (f); f\in R,\ \nu_2(f)=\phi_2\}$. Note that $\lambda(\tilde \phi_2)=\phi_2$ and that $t_1\in \hbox{\rm Ker}\lambda$.\par
We remark that for $\phi_1\in [\tilde \phi_2,\tilde\phi_2+y_1t_1]$ we have the inclusions\par\noindent
$$p_1^{y_1}\P_{\phi_2}\subset \P_{\phi_1}\subset \P_{\phi_2},$$ 
$\P_{\phi_2}=\P_{\tilde \phi_2}$ and since the valuation $\overline\nu_1$ which is the image of $\nu_1$ in $R/p_2$ is of rank one, the number of elements in the interval $ [\tilde \phi_2,\tilde\phi_2+y_1t_1]$ is finite.\par
It follows that for any additive function $A$ on $R$-modules we have 
$$
\sum_{\phi_1\in[\tilde \phi_2,\tilde\phi_2+y_1t_1]}A(\P_{\phi_1}/\P_{\phi_1}^+)\leq A(M_{\phi_2}/p_1^{y_1}M_{\phi_2}),\leqno(1)
$$ 
where $M_{\phi_2}= \P_{\phi_2}/ \P^+_{\phi_2},$ a finitely generated torsion free $R/p_2$-module.\par\noindent\par\noindent
The $R$ modules $\P_{\phi_1}/\P_{\phi_1}^+$ and $M_{\phi_2}/p_1^{y_1}M_{\phi_2}$ have $p_1$ as their unique minimal prime.

In the special case where $\nu_1$ is a valuation of rank one and $\nu_2$ is the trivial valuation, we have $\Gamma_2=0$, $\tilde \phi_2=0\in \Gamma_1$ and, writing $\phi$ for $\phi_1$, $p$ for $p_1$, etc.,  the inequality reduces to:
$$\sum_{\phi\in[0,yt]}A(\P_{\phi}/\P_{\phi}^+)\leq A(R/p_1^{y}R),\leqno(2)$$

\par\noindent
We can take as additive function $A$ on $R$-modules of dimension $=\hbox{\rm dim}R/p_1$ the multiplicity $e_m$  with respect to the maximal ideal of $R$.  The inequality $(1)$ then becomes 

$$\sum_{\phi_1\in[\tilde \phi_2,\tilde\phi_2+y_1t_1]}e_m(\P_{\phi_1}/\P_{\phi_1}^+)\leq e_m(M_{\phi_2}/p_1^{y_1}M_{\phi_2}).\leqno(3)$$
It is now possible to evaluate the right hand side thanks to ([B], \S7, No.1, Prop 3), which says this: given a finitely generated $R$-module $M$ of dimension $d\geq 0$, denote by $B$ the set of minimal prime ideals of $\hbox{\rm Supp}(M)$ such that $\hbox{\rm dim}(R/p)=d$. Let $q$ be an ideal of $R$ contained in its radical and such that $M/qM$ has finite length. Then we have 
$$e_q(M)=\sum_{p\in B}\hbox{\rm length}_{R_p}  (M_p)e_q(R/p).$$ If we apply this to $M_{\phi_2}/p_1^{y_1}M_{\phi_2}$ we get:\par\noindent
$$
e_m(M_{\phi_2}/p_1^{y_1}M_{\phi_2})= \hbox{\rm length}_{R_{p_1}}\bigl((M_{\phi_2})_{p_1}/m^{y_1}_1(M_{\phi_2})_{p_1}\bigr)e_m(R/p_1),
\leqno(4)
$$
where $m_1=p_1R_{p_1}$ and the support of the $R$-module $M_{\phi_2}$ is the ideal $p_2$.\par\noindent
By the main result on Hilbert-Samuel polynomials, in view of the fact that the support of $(M_{\phi_2})_{p_1}$ is the ideal $p_2R_{p_1}$, we have an asymptotic estimate:
$$\hbox{\rm length}_{R_{p_1}}\bigl((M_{\phi_2})_{p_1}/m^{y_1}_1(M_{\phi_2})_{p_1}\bigr)\asymp \frac{e_{m_1}\bigl((M_{\phi_2})_{p_1})\bigr)}{(\hbox{\rm dim}(R_{p_1}/p_2R_{p_1})!}y_1^{\hbox{\rm dim}(R_{p_1}/p_2R_{p_1})}.\leqno(5)$$
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Let $$(0)=p_{n+1}\subset p_n\subset p_{n-1}\subset \cdots \subset p_1$$
be the sequence of the centers of valuations $\nu_i$ with which a given valuation $\nu$ of rank $n$ on a noetherian local domain $(R,m)$ is composed. We assume that the centers are distinct prime ideals.

Let $p_0$ be a prime ideal of $R$ containing the center $p_1$ of $\nu=
\nu_1$ ($p_0$ could be equal to $p_1$).
Then
the sum
$$\sum_{\phi_n\in [0,t_ny_n]}\sum_{\phi_{n-1}\in [
\tilde\phi_n,\tilde\phi_n+t_{n-1}y_{n-1}]}\cdots \sum_{\phi_1\in [
\tilde\phi_{2},\tilde\phi_{2}+t_1y_1]}e_{m_0}((\P_{\phi_1}/\P^+_{\phi_1})_{p_0})$$
is bounded for $y_1,\ldots ,y_n$ large enough by a function which behaves asymptotically as
$$\frac{\Pi_{i=0}^n e_{m_i}((R/p_{i+1})_{p_i})}{\Pi_{i=1}^n(\hbox{\rm dim}(R/p_{i+1})_{p_i})!} \Pi_{i=1}^ny_i^{\hbox{\rm dim }(R/p_{i+1})_{p_i}},$$
where $m_i=p_iR_{p_i}$ for $0\le i\le n$. We  take
$y_1\gg y_2\gg \cdots\gg y_n$ to obtain this asymptotic behavior.\par\noindent

The proof of this formula is by induction on $n$. We first prove the formula in the case when $n=1$. We apply formulas $(2)$ and $(3)$ - $(5)$ to the ring $R_{p_0}$, observe that for 
$\phi_1\in\Gamma_1$,
$$
(\P_{\phi_1})_{p_0}=\{f\in R_{p_0}\mid \nu(f)\ge \phi_1\},
$$
and that $(R_{p_0})_{p_1R_{p_0}}\cong R_{p_1}$, to obtain
$$
\sum_{\phi_1\in [0,t_1y_1]}e_{m_0}((\P_{\phi_1}/\P_{\phi_1}^+)_{p_0})
\asymp
\frac{e_{m_1}(R_{p_1})e_{m_0}((R/p_1)_{p_0})}{(\mbox{dim }R_{p_1})!}y_1^{\mbox{dim }R_{p_1}},
$$
which is the formula for $n=1$.

We now assume that the formula is true for valuations of rank $<n$. We will derive the formula for a rank $n$ valuation $\nu$.   We apply the formula to the rank $n-1$ valuation $\nu_2$ which $\nu$ is composite with, and the chain of prime ideals
$$
(0)=q_n\subset q_{n-1}\subset \cdots\subset q_1\subset q_0
$$
where $q_{n-1}=p_{n},\ldots,q_1=p_2$ are the centers on $R$ of the successive valuations
$\nu_n,\ldots,\nu_2$ with which $\nu_2$ is composite, and   $q_0=p_1$, to obtain
$$
\sum_{\phi_n\in[0,t_ny_n]}\cdots\sum_{\phi_2\in [\tilde \phi_3+t_2y_2]}
e_{m_1}((\P_{\phi_2}/\P_{\phi_2^+})_{p_1})\asymp
\frac{\prod_{i=1}^ne_{m_i}((R/p_{i+1})_{p_i})}
{\prod_{i=2}^n(\mbox{dim }(R/p_{i+1})_{p_i})!}
\prod_{i=2}^ny_i^{\mbox{dim }(R/p_{i+1})_{p_i}}.\leqno(6)
$$
We now apply formulas $(1)$, and $(3)-(5)$ to the ring $R_{p_0}$ and the valuation $\nu=\nu_1$, to obtain
$$
\sum_{\phi_1\in[\tilde\phi_2,\tilde\phi_2+y_1t_1]}e_{m_0}((\P_{\phi_1}/\P_{\phi_1}^+)_{p_0})
\asymp
\frac{e_{m_1}((\P_{\phi_2}/\P_{\phi_2}^+)_{p_1})e_{m_0}((R/p_1)_{p_0})}
{(\mbox{dim }(R/p_2)_{p_1})!}
y_1^{\mbox{dim }(R/p_2)_{p_1}}.\leqno(7)
$$
Finally, we  sum over $(7)$ and $(6)$ to obtain the desired formula for $\nu$.

\end{document}
The proof is by descending induction on $j$. The first step of the induction concerns the valuation $\nu_n$, which is of rank one and is given by
%-----------
\par\noindent
Define $ t_j=  \nu_j(p_j)\in \Gamma_j$ for each $j$ and remark that it is in the kernel of the map $\Gamma_j\to \Gamma_{j+1}$, which is of rank one. We can therefore think of it as a real number.\par\noindent 
Lemma: For each $j$ we have $p_j^{y_j}\P_{\phi_{j-1}}\subset \P_{\tilde\phi_{j-1}+t_jy_j}$ and the interval $[\tilde\phi_{j-1},\tilde\phi_{j-1}+t_jy_j]$ is finite.\par\noindent

Assume that the result is known for $j$ and prove it for $j-1$. Let us apply the result above to \par
In view of the lemma, we have the inequality
$$\sum_{\phi_j\in [\tilde\phi_{j+1},\tilde\phi_{j+1}+t_jy_j]}e_m(\P_{\phi_j}/\P_{\phi_j}^+)\leq e_m(M_{\phi_{j+1}}/p_j^{y_j}M_{\phi_{j+1}}).$$
Now we apply the formula ([B], \S7, No.1, Prop 3) $$e_m(M_{\phi_{j+1}}/p_j^{y_j}M_{\phi_{j+1}})\asymp \frac{e_{m_j}((M_{\phi_{j+1}})_{p_j})}{(\hbox{\rm dim}R_{p_j}/p_{j-1}R_{p_j})!}y_j^{\hbox{\rm dim}R_{p_j}/p_{j-1}R_{p_j}},$$
so that for each $\phi_{j+1}\in \Gamma_{j+1}$ we have for any $\epsilon>0$ for $y_j$ sufficiently large (in a way which depends on $\epsilon$ and $\phi_j$) an inequality
$$\sum_{\phi_j\in [\tilde\phi_{j+1},\tilde\phi_{j+1}+t_jy_j]}e_m(\P_{\phi_{j+1}}/\P_{\phi_j}^+)\leq (1+\epsilon)\frac{e_{m_j}(M_{\phi_j)}}{(\hbox{\rm dim}R_{p_j}/p_{j-1}R_{p_j})!}y_j^{\hbox{\rm dim}R_{p_j}/p_{j-1}R_{p_j}} .$$

Summing up over the $\phi_{j+1}\in [\tilde\phi_{j+2},\tilde\phi_{j+2}+t_{j+1}y_{j+1}] $, which are finite in number, then over the $\phi_{j+2}\in [\tilde\phi_{j+3},\tilde\phi_{j+3}+t_{j+2}y_{j+2}] $ and so on, we obtain the statement.
  

\end{document}

Since $\nu_1$ is a valuation of rank one, we have that $p_1^{y_1}\subset \P_{\phi_1}$ provided that $\phi_1\in [0, t_1y_1]$, and the number of elements in that interval is finite.\par
Now if $A$ is such that $ A(M_{\phi_1}/m^{y_0}M_{\phi_1})$ is a polynomial in $y_0$ for large $y_0$, it is asymptotically dominated by an expression $A_0(M_{\phi_1})y_0^{d_1}$, where $A_0$ is additive on modules under dimension restrictions.
We see that 
$$\sum_{\phi_1\in [0,y_1t_1]}A_0(M_{\phi_1})=A_0(R/\P_{y_1t_1})\leq A_0(R/p_1^{y_1})$$
Now we have a formula for the multiplicity 
$$e_m(R/p_1^{y_1})=\hbox{\rm length}_{R_1}(R_1/m_1^{y_1})e_m(R/p_1)$$
where $R_1=R_{p_1}$ and $m_1=p_1R_1$.\par
If we write out the Hilbert-Samuel function of $R_1$ as
$$\hbox{\rm length}_{R_1}(R_1/m_1^{y_1})=e_{m_1}(R_1)\frac{y_1^{\hbox{\rm ht}p_1}}{(\hbox{\rm ht}p_1)!}+\cdots,$$ we find that we can write, taking for $A$ the $m$-adic multiplicity and adding up the inequalities (1) for $\phi_1\in[0,t_1y_1]$:
$$\sum_{\phi_1\in [0,y_1t_1]}\sum_{\phi\in[\tilde \phi_1,\tilde\phi_1+y_0t_0]}e_m(\P_\phi/\P_\phi^+)\prec \frac{e_m(R/p_1)}{(\hbox{\rm dim}R/p_1)!}\frac{e_{m_1}(R_1)}{(\hbox{\rm ht}p_1)!}y_0^{\hbox{\rm dim}R/p_1}y_1^{\hbox{\rm ht}p_1}$$ where $a\prec b$ means  $a\leq (1+\epsilon )b$ for any $\epsilon >0$ and $t_0, t_1$ large enough.
 