The homogenized ideal is equivalent to the marked ideal

Equality of supports at every stage. Fix a multiple test blow-up $(X_k)$ of $(\mathcal{I},\mu )$. By [F4], each derivative factor and the tangent ideal have controlled transforms along this same sequence. At every stage, [F2] gives $\mathrm{supp}({\mathcal{I}}_k,\mu )\subseteq \mathrm{supp}({\mathcal{D}}^i({\mathcal{I}}_k),\mu -i)$ for $i<\mu$, and also $\mathrm{supp}({\mathcal{I}}_k,\mu )\subseteq \mathrm{supp}(T({\mathcal{I}}_k),1)$. By [F4], the actual transformed derivative and tangent factors are contained in the corresponding derivative ideals of ${\mathcal{I}}_k$; their product is thus contained in ${\mathcal{D}}^i({\mathcal{I}}_k)T({\mathcal{I}}_k{)}^i$. The product order inequality therefore puts every point of $\mathrm{supp}({\mathcal{I}}_k,\mu )$ in the support of each transformed summand $({\mathcal{D}}^i({\mathcal{I}}_k)T({\mathcal{I}}_k{)}^i,\mu )$. The support of the literal sum is the intersection of the summand supports: for ideals $J_i$, ${\mathrm{ord}}_x({\sum }_i{J}_i)={\min }_i{\mathrm{ord}}_x(J_i)$. Hence $\mathrm{supp}({\mathcal{I}}_k,\mu )\subseteq \mathrm{supp}(H(\mathcal{I},\mu {)}_k)$. Conversely, the first summand of $H$ is $\mathcal{I}$, so ${\mathcal{I}}_k\subseteq H(\mathcal{I},\mu {)}_k$ and $\mathrm{supp}(H(\mathcal{I},\mu {)}_k)\subseteq \mathrm{supp}({\mathcal{I}}_k,\mu )$. This proves equality of supports at every stage.

Equivalence and transform decomposition. Equality of supports at every stage in step 1.1 shows that the two marked ideals have the same admissible centers and the same induced supports along every multiple test blow-up. By [F5] they are equivalent, proving (1). For (2), the controlled transform of the literal ideal sum defining $H(\mathcal{I})$ distributes termwise because all summands have mark $\mu$; each transformed summand is the product of the transforms of $({\mathcal{D}}^i(\mathcal{I}),\mu -i)$ and $(T(\mathcal{I}),1{)}^i$ by [F3, F4]. Thus its underlying ideal with mark $\mu$ is the literal sum displayed in the Statement. Under AC, the iterated marked sum has the intersection of the transformed summand supports and exactly their simultaneous test sequences by [F3]. The literal sum with common mark $\mu$ has that same support, and its transforms continue to distribute termwise at every common admissible center. Induction gives equal induced supports and the same test sequences, hence equivalence by [F5], proving (2).