The support is the union of the closures of the associated primes

By [L3], write ${\mathrm{Supp}}_R(M)=V({\mathfrak{p}}_1)\cup \cdots \cup V({\mathfrak{p}}_n)$ for the prime ideals ${\mathfrak{p}}_i$ occurring in a prime filtration of $M$. Let $\mathfrak{q}\in {\mathrm{Supp}}_R(M)$. Then ${\mathfrak{p}}_i\subseteq \mathfrak{q}$ for some $i$. Choose such an index $i$ with ${\mathfrak{p}}_i$ minimal under inclusion among the filtration primes contained in $\mathfrak{q}$. If $\mathfrak{r}\in {\mathrm{Supp}}_R(M)$ and $\mathfrak{r}\subseteq {\mathfrak{p}}_i$, then [L3] gives some ${\mathfrak{p}}_j\subseteq \mathfrak{r}\subseteq {\mathfrak{p}}_i$, so the minimal choice of ${\mathfrak{p}}_i$ forces ${\mathfrak{p}}_j={\mathfrak{p}}_i$ and hence $\mathfrak{r}={\mathfrak{p}}_i$. Therefore ${\mathfrak{p}}_i$ is minimal in the support, so [L1] gives ${\mathfrak{p}}_i\in {\mathrm{Ass}}_R(M)$ and $\mathfrak{q}\in V({\mathfrak{p}}_i)$. This proves ${\mathrm{Supp}}_R(M)\subseteq {\bigcup }_{\mathfrak{p}\in {\mathrm{Ass}}_R(M)}V(\mathfrak{p})$.

Conversely, let $\mathfrak{p}\in {\mathrm{Ass}}_R(M)$ and let $\mathfrak{q}\supseteq \mathfrak{p}$. By [L4], the prime $\mathfrak{p}$ lies in the support; then [L2] gives ${\mathrm{Ann}}_R(M)\subseteq \mathfrak{p}\subseteq \mathfrak{q}$, so $\mathfrak{q}\in {\mathrm{Supp}}_R(M)$. Thus $V(\mathfrak{p})\subseteq {\mathrm{Supp}}_R(M)$ for every $\mathfrak{p}\in {\mathrm{Ass}}_R(M)$.

Steps 1.1 and 1.2 prove the support decomposition.