If $\Phi$ is absolutely continuous and singular then $\Phi$ is null

Question

Definition

1. $\Phi$ is concentrated in a measurable set $A$ when $\Phi(E) = 0$ for each $E \subset X-A$, with $E$ belongs to ring $\mathbb{G}$.
2. $\Phi$ is absolutely continuous (with respect to measure $\mu$) when $\Phi(A) = 0$ for all measurable $A$ such that $\mu(A) = 0$.
3. $\Phi$ is called singular (with respect to the $\mu$ measure) when is concentrated in a set of $\mu$-null measure

Problem If $\Phi$ is absolutely continuous and singular then $\Phi$ is null.

Attempt Let $\Phi$ singular in $A$ then $\Phi(E)=0$ for all $E \subset X-A$ with $\mu(A)=0$. But $\Phi$ is absolutly continuous we have $\mu(E)=0$ for all $E \subset X-A$. But I don't see more, or did I finish?. Could you guide me?

Answer 1

Let $\Phi$ be concentrated in the set $A$ with $\mu(A)=0.$ Then since $X-A\subset X-A,$ we have by definition (1) that $\Phi(X-A)=0$. By definnition of absolute continuity, we also have $\Phi(A)=0.$ Hence, we have $\Phi(X) = \Phi(X-A) + \Phi(A) = 0.$ And hence for any measurable set $E\subset X$ we have $\Phi(E)\leq \Phi(X) =0$.