The $\Omega$-Stability Theorem

Question

I'm currently studying the $\Omega$-Stability Theorem:

Theorem: If $\mathcal{R}(f)$ has a hyperbolic structure then $f$ is $\mathcal{R}$-stable.

Some explanations about the statement: $f$ is a $C^1$ diffeomorphism of a compact manifold; $\mathcal{R}(f)$ denotes the chain recurrent set of $f$; $\mathcal{R}$-stable means that there is a $C^1$ neighborhood of $ f $, $\mathcal{N}(f)$, such that for every $g\in \mathcal{N}(f)$ we have that $g|\mathcal{R}(g)$ is conjugated to $f|\mathcal{R}(f).$

Some explanations about the proof of the theorem:

Fact 1: If $\Lambda_f$ is a hyperbolic set for $f$ with local structure product then there exists a $C^1$ neighborhood, $\mathcal{N}(f)$, of $f$ and a open $U\supset \Lambda$ such that for every $g\in \mathcal{N}(f)$ we have that $f|\Lambda_f$ is conjugated to $g|\Lambda_g$, where $\Lambda_g=\bigcap_{n\in \mathbb{Z}}g^n(U)$.

Proof: The stability of the hyperbolicity follows from this answer. The conjugation between $f$ and $g$ follows by observing that the zero section $0_{\Lambda}$ is a hyperbolic fixed point o the operator $F:\Gamma^{0}(T_{\Lambda}M)(\epsilon)\to \Gamma^{0}(T_{\Lambda}M)(\epsilon)$ given by $$ F(\gamma)(x):=\exp_{x}^{-1}f\exp_{f^{-1}(x)}\gamma(f^{-1}(x)) $$ where $\epsilon$ is is the injectivity radius of $\exp$.

Now observe that there is a conjugation $h$ between $f$ and $g$ if and only if the section $\gamma(x):=\exp_{x}^{-1}h(x)$ is a fixed point for the operator $$ F^{g}(\gamma)(x):=\exp_{x}^{-1}g\exp_{f^{-1}(x)}\gamma(f^{-1}(x)) $$ so if we take $ g $ sufficiently close to $f$, the $F^g$ operator will be a small perturbation of the operator $F$, so close to the null section should be a fixed point for the operator $F^g$.

Fact 2: There is a filtration for $f$, i.e. a family of manifolds $\{M_{j}\}_{0\leq j\leq N}$ of $0$ codimension such that:

1. $M=M_N\supset M_{N-1}\supset M_{N-2}\supset\cdots \supset M_{0}=\varnothing$
2. $f(M_i)\subset int(M_i)$
3. $\Lambda_j\subset int(M_i\setminus M_{i-1})$ and $\Lambda_j=\bigcap_{n\in\mathbb{Z}}f^{n}(M_i\setminus M_{i-1}).$

So given this filtration and using the fact 1, isolating neighborhoods $U_{i}= M_{j}\setminus M_{j-1}$, and a neighborhood of $f$, $\mathcal{N}$ s.t for $g\in \mathcal{N}$ we have:

1. Each of the $U_{i}=M_{j}\setminus M_{j-1}$ is a isolating neighborhood for $\Lambda_{j}^{g}:=\bigcap_{n\in\mathbb{Z}}g^{n}(M_i\setminus M_{i-1})$
2. There is $h_{j}:\Lambda_{j}^{f}\to \Lambda_{j}^{g}$ that is a conjugacy
3. $g(M_{j})\subset int(M_{j})$

So we define $h:\bigcup_{j=1}^{N}\Lambda_{j}^{f}\to \bigcup_{j=1}^{N}\Lambda_{j}^{g}$, by $h|\Lambda_{j}^f=h_j$. This gives a conjugacy on these sets.

Fact3: $\mathcal{R}(g)=\bigcup_{j=1}^{N}\Lambda_{j}^{g}$

This is a simple exercise using the Conley Theorem: Conley Theorem (or fundamental theorem of dynamical systems)

My question: In order to demonstrate the fact 2, Robinson uses the following argument. There is only a finite number of classes by recurrence chains that currently match the basic sets of $f$ (this follows from the Spectral theorem): see Relationship between basic sets and attractors. It follows from the no cycle property that each pairs $\Lambda_{j}$ and $\Lambda_{k}$ can be put in different attracting-repelling pairs so the Liapunov function $L:M\to \mathbb{R}$ has different values on each of the $\Lambda_j$. So $L$ can be modified so that $L(\Lambda_j)=j$. Using this modified function, we define $$ M_j=L^{-1}(-\infty,j+\frac{1}{2}] $$ So my questions is how to modify the Liapunov function $L$ in order to have the above properties?