If $x\in I$ is such that $\text{Supp}(0:_M x)\subseteq V(I)$ , then how to show that $x \notin \cup_{P\in \text{Ass}_R(M)\setminus V(I)} P $? Let $M$ be a finitely generated module over a Noetherian ring $R$. Let $I$ be an ideal of $R$. Since Ass$_R (M)$ is finite, and $I\nsubseteq P$ for all $P\in$Ass$_R (M)\setminus V(I)$, so by prime avoidance $I\nsubseteq \cup_{P\in \text{Ass}_R(M)\setminus V(I)} P$.
Now let $x\in I\setminus \cup_{P\in \text{Ass}_R(M)\setminus V(I)} P$. If $P\in \text{Ass}(0:_M x)$ then $P \in \text{Ass}(M)$ and $P=\text{ann}_R(m)$ for some $m\in (0:_M x)$. Hence $x\in \text{ann}_R(m)=P$. So by choice of $x$, we have $P\in V(I)$. Thus we have $\text{Ass}(0:_M x) \subseteq V(I)$, hence $\text{Supp}(0:_M x)\subseteq V(I)$.
My question is how to show the converse: If $x\in I$ is such that
$\text{Supp}(0:_M x)\subseteq V(I)$ , then how to show that
$x \notin \cup_{P\in \text{Ass}_R(M)\setminus V(I)} P$?
(Here $(0:_M x):=\{m\in M: xm=0\}$)
Please help. Thanks.

Since $Supp(0:_M x)=V(Ann(0:_M x))\subseteq V(I)$, this implies that $rad(I)\subset rad(Ann(0:_M x)). $
Hence, since $R$ is noetherian, $I^n\subseteq Ann(0:_M x)$ for some $n>0$.
Let $x\in P$ for some prime ideal $P \in Ass_R(M)-V(I)$. So $P=Ann_R(m)$ for some $0\neq m\in M$. In particular, $xm=0$.
Now by discussion in the first paragraph, $Ann(0:_M x)\nsubseteq P$. Let $y\in  Ann(0:_M x)$ such that $y\notin P$.
Now $0\neq m\in (0:_M x). $ So $ym=0$, hence $y\in Ann_R(m)=P$, a contradiction to the fact that $y\in P$.
So we are done.
