Suppose that $\mu$ is a measure on a measurable set $(X,\mathcal{B})$. And suppose that $A\subseteq X$ is not in $\mathcal{B}$.
Is there a measure $\mu'$ on the $\sigma$-algebra generated by $\mathcal{B}\cup \{A\}$ that coincide with $\mu$ on $\mathcal{B}$ and is such that $\mu'(A)=0$ ?
Borrowing from the above link:
Let $a := \sup\{\mu(B) : B \subseteq A, B \in \mathcal{B}\}$. There exists $K$ satisfying $K \subseteq A$ and $K \in \mathcal{B}$ and $\mu(K) = a$, (namely, $K := \bigcup_{B \in \mathcal{B} : B \subseteq A} B)$.
Note that the new $\sigma$-algebra can be expressed as $\{(B_1 \cap A) \cup (B_2 \cap A^c) : B_1, B_2 \in \mathcal{B}\}$.
We define $\mu'((B_1 \cap A) \cup (B_2 \cap A^c)) := \mu(B_1 \cap K) + \mu(B_2 \cap K^c)$. One can check that $\mu'$ is a measure, that $\mu'(B) = \mu(B)$ for $B \in \mathcal{B}$, and that $\mu'(A) = \mu(K) = a$. So if $a$ (as defined above) is zero (which I believe is equivalent to the condition in d.k.o.'s comment above), then this construction gives you what you want.