I have a question about F. Bastianelli's article: Measure of irrationality for hypersurfaces of large degree
We have the Definition 1.1.
A line bundle $L$ on $X$ satisfies property $(BVA)_p$ if there exists a proper Zariski-closed subset $Z = Z(L) \subset X$ depending on L such that $$ H^0(X,L) \to H^0(X, L \otimes \mathcal{O}_{\xi}) $$ surjects for every finite subscheme $\xi \subset X$ of length $p+1$ whose support is disjoint from $Z$.
Thus $(BVA)_0$ is equivalent to requiring that $L$ be effective, and $(BVA)_1$ is what is often called "birationally very ample".
My question
I wanted to show that if $L$ is a line bundle on $X$ satisfying $(BVA)_p$ and $E$ is an effective divisor on $X$, then $\mathcal{O}_X(L+E)$ satisfies $(BVA)_p$.
What I did so far
By applying the definition above, since $L$ satisfies $(BVA)_p$ then there exist a Zariski-closed subset $Z$ such as for every finite subscheme $\xi \subset X$ of length $p+1$ whose support is disjoint from $Z$, the map $$ H^0(X,L) \to H^0(X, L \otimes \mathcal{O}_{\xi}) $$ is surjective.
On the other hand, since $E$ is effective, then it verifies $(BVA)_0$ (i.e. there exist a Zariski-closed subset $Z'$ such as for every finite subscheme $\xi' \subset X$ of length $1$ whose support is disjoint from $Z'$, the map $$ H^0(X,E) \to H^0(X, E \otimes \mathcal{O}_{\xi'}) $$ is surjective.)
Where I'm stuck
To show that $\mathcal{O}_X(L+E)$ satisfies $(BVA)_p$, which Zariski-closed subset that we'll take? My suggestion is $Z \cup Z'$, am I right? and if the answer is yes, can I take any subscheme $\Xi \subset X \backslash (Z \cup Z')$ for the finite subscheme of length $p+1$ whose support is disjoint from $Z \cup Z'$ ?
Thanks in advance for any hint or enlightement?
K. Y.
The answer is yes.
It is enough to see that since $E$ is effective, then $L + E_{|X \ Supp E} = L \otimes \mathcal{O}_X = L$.