Suppose that H is generated by H0∪H1 and M generated by ⨁i≤n0Mi for a suitable integer n0. Then the filtration (S−1M≥n) on S−1M is good for the ideal S−1H≥1 of S−1H.
One has, for n≥n0, M≥n+1=H1⋅M≥n, hence S−1M≥n+1=H1⋅S−1M≥n=S−1H≥1⋅S−1M≥n.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.