Filtration, asymptotic $sigma$-prime divisors and superficial elements

Let $(A,mathfrak{M})$ be a Noetherian local ring with infinite residue field $A/ mathfrak{M}$ and $I$ be a $mathfrak{M}$-primary ideal of $A$. Let $f = (I_{n})_{nin mathbb{N}}$ be a good filtration on $A$ such that $I_{1}$ containing $I$. Let $sigma$ be a semi-prime operation in the set of ideals of $A$. Let $lgeq 1$ be an integer and $(f^{(l)})_{sigma} = sigma(I_{n+l}):sigma(I_{n})$ for all large integers $n$ and$rho^{f}_{sigma}(A)= min big{ nin mathbb{N} | sigma(I_{l})=(f^{(l)})_{sigma}, for all lgeq n big}$. Here we show that, if $I$ contains an $sigma(f)$-superficial element, then $sigma(I_{l+1}):I_{1}=sigma(I_{l})$ for all $l geq rho^{f}_{sigma}(A)$. We suppose that $P$ is a prime ideal of $A$ and there exists a semi-prime operation $widehat{sigma}_{P}$ in the set of ideals of $A_{P}$ such that $widehat{sigma}_{P}(JA_{P})=sigma(J)A_{P}$, for all ideal $J$ of $A$. Hence $Ass_{A}big( A / sigma(I_{l}) big) subseteq Ass_{A}big( A / sigma(I_{l+1}) big)$, for all $l geq rho^{f}_{sigma}(A)$.