b_iB_i$ where $b_i\in \mathbb R$ and $B_i$ is a prime divisor on $V$ for every $i$ with $B_i\ne B_j$ for $i\ne j$. Assume that $A\sim _{\mathbb R} B$. Then we can take a $\mathbb Q$-Cartier $\mathbb Q$-divisor $C=\sum _{i\in I} c_i B_i$ on $V$ such that

-   $A\sim _{\mathbb Q} C$,

-   $c_i=b_i$ holds if $b_i\in \mathbb Q$, and

-   $|c_i-b_i|\ll 1$ holds for $b_i\in \mathbb R\setminus 
    \mathbb Q$.

In particular, ${{\operatorname{Supp}}}C={{\operatorname{Supp}}}B$, $\lfloor C\rfloor 
=\lfloor B\rfloor$, $\lceil C\rceil =\lceil B\rceil$, and ${{\operatorname{Supp}}}\{C\}={{\operatorname{Supp}}}\{B\}$.

It is an easy exercise. For the details, see, for example, the proof of [@fujino-fundamental Lemma 4.15].

\[x-lem3.2\] Let $D=\sum _{i\in I} a_i D_i$ be an $\mathbb R$-divisor on a smooth projective variety $V$, where $a_i\in \mathbb R$ and $D_i$ is a prime divisor on $V$ for every $i$ with $D_i\ne D_j$ for $i\ne j$. Assume that $D$ is semi-ample. Then we can construct a $\mathbb Q$-divisor $D^\dag=\sum _{i\in I} b_iD_i$ such that

-   $D^\dag$ is semi-ample,

-   $b_i=a_i$ holds if $a_i \in \mathbb Q$, and

-   $|b_i-a_i|\ll1$ holds for $a_i\in \mathbb R\setminus \mathbb Q$.

Since $D$ is semi-ample, we can write $D=\sum _{j\in J} m_jM_j$ where $m_j\in \mathbb R$ and $M_j$ is a semi-ample Cartier divisor on 