Let $S = k[x_0,x_1,\ldots,x_n]$ with its usual grading and let $I \subset S$ be a homogeneous ideal not containing $S_+ = (x_0,x_1,\ldots,x_n)$. We define the saturation of $I$ to be the homogeneous ideal \begin{equation} I^{\text{sat}} = \{f \in S \mid f \cdot S_n \subseteq I \text{ for some } n\} \supseteq I \end{equation}
I want to show that for $d$ sufficiently large, we have $I^{\text{sat}}_d = I_d$.
I know this should be true since the saturated ideal defines the same projective subvariety as $I$, and isomorphic varieties have the same Hilbert polynomial, so $(S/I)_d = (S/I^{\text{sat}})_d$ for $d \gg 0$. However, I'm looking for a more straightforward proof.
$I^{\text{sat}}$ is finitely generated, so there is $r\ge 1$ such that $S_rI^{\text{sat}}\subseteq I$. Since $S_{r+i}=S_iS_r$ we get $S_kI^{\text{sat}}\subseteq I$ for all $k\ge r$. This shows that $S_+^rI^{\text{sat}}\subseteq I$, and thus the graded $S$-module $I^{\text{sat}}/I$ is artinian (as being a finitely generated module over the artinian ring $S/S_+^r$). But graded artinian modules have all graded components equal to zero from some degree on.