M is an R-module which is not finitely generated. is it true that $\inf \{ i| H^i_I(M)\neq 0 \}\le ht_M I?$ Let $R$ be a commutative noetherian ring (with $1$), $I$ an ideal of $R$, and $M$, a finitely generated $R$-module such that $IM \neq M$. Then, by Theorem 6.2.7 of BRODMANN-SHARP's Local Cohomology book one has
$$\underline{ \inf \{ i| H^i_I(M)\neq 0 \}} \color{silver}{(=grade (I,M))}\underline{ \le ht_M\ I}.$$ 
Here  $ht_M\ I= \inf \{ ht_M\ q| q\in supp M\cap V(I)\},$ where, $ht_M\ q$ is the Krull dimension of the $R_q$-module $M_q$.  
What happens (to $\le$), if we drop the assumption "$M$, is a finitely generated $R$-module"?   
What if we replace "$M$, is a finitely generated $R$-module" by "$M$, is an $R$-module of dimension $0$ "? (Here, Also we can take $I$, prime)
 A: For the concept of the grade of an arbitrary module with respect to a finitely generated ideal see the section 9.1 (entitled as "Grade and acyclicity") of the book entitled as "Cohen-Macaulay Rings" written by Herzog and Bruns. In particular, see the Proposition 9.1.2. When $R$ is Noetherian the local cohomology is just the direct limit of Koszul homologoies (See, Theorem 5.2.9.  of the book "Local Cohomology" written by Sharp and Brodmann). So, Definitely, their definition, directly, yields $$\text{inf}\{i;\text{H}^i_I(M)\neq 0\}\ge \text{grade}(I,M).$$ Then a standard inductive argument on the grade shows that the equality holds, i.e., $$\text{inf}\{i;\text{H}^i_I(M)\neq 0\}=\text{grade}(I,M).$$ 
If $M$ has zero dimension then the support of $M$ is a subset of the set of maximal ideals of of $R$. Let, $\mathfrak{m}\in \text{Supp}(M/IM)$. Then, $\mathfrak{m}\in \text{Supp}(M).$ Hence there exists $x\in M$ such that $0:_Rx\subseteq \mathfrak{m}$. Set, $N=Rx$. Then since $\text{Supp}(Rx)\subseteq \text{Supp}(M)$ so we have $\dim(Rx)=0$. Furthermore, $\mathfrak{m}R_\mathfrak{m}\in \text{Ass}_{R_\mathfrak{m}}(R_{\mathfrak{m}}x/1)$ so $\mathfrak{m}\in \text{Ass}(Rx)$. This shows that $\Gamma_I(M)\neq 0$.
$ \ \ \ $For the inequality of $\text{ht}_M(I)$: Claim: $\text{grade}(I,M)\le\text{ht}_M(I)$: 
  If $\text{ht}_M(I)=\infty$ then we are done. So let $\mathfrak{p}\in\text{Var}(I)\bigcap \text{Supp}_R(M)$ with $h=\text{ht}_M(I)=\text{dim}(M_\mathfrak{p})$. Then it suffices to prove that $\text{grade}(IR_\mathfrak{p},M_\mathfrak{p})\le \text{dim}_{R_\mathfrak{p}}(M_\mathfrak{p})$. So we may assume that $(R,\mathfrak{m})$ is local and then prove that $\text{grade}(I,M)\le \text{dim}(M)$. Assuming the contrary  we get $H^i_{I}(M)\neq 0$ for  $i=\text{grade}(I,M)\gneq \text{dim}(M)$ (by the above formula). This contradicts with the Grothendieck's vanishing theorem (This theorem, as stated in the Sharp and Brodmann's book,  is not restricted to the category of  finitely generated modules).
