Let $\mu$ be a measure and $\mu_1, \mu_2$ are mutually singular measures on a space $(X, \mathcal{B})$ where $\mu = \mu_1 - \mu_2$. Show $\mu_2 = 0$.

Question

Here is my proof. Looking for:

1. Proof verification - is this correct? Most important!
2. Suggestions on ways to make it better if it is correct. Secondary
3. Other proofs. Curious. Is there another approach to this proof completely different from the one I posted.

Thanks!

Proof: Let $\mu$ be a measure and $\mu_1$ and $\mu_2$ be mutually singular measures on a space $(X, \mathcal{B})$. We show if $\mu = \mu_1 - \mu_2$ then $\mu_2 = 0$. Since $\mu_1$ and $\mu_2$ are mutually singular we can consider $X$ a disjoint union of measurable sets $A$ and $B$ such that $X = A \cup B$ and

$$\mu_1(B) = 0 = \mu_2(A).$$

Assume $\mu = \mu_1 - \mu_2$. Since $\mu$, $\mu_1$, and $\mu_2$ are all measures by definition they are all greater than or equal to zero. Now assume $E \subset B$ and consider,

$$\mu(E) = \mu_1(E) - \mu_2(E).$$

then $\mu_1(E) = 0$ so we would have

$$\mu(E) = 0 - \mu_2(E).$$

This contradicts the fact that $\mu(E) \ge 0$ and thus $\mu_2(E) = 0$. Therefore if $\mu$ is a measure and $\mu_1$ and $\mu_2$ are mutually singular measures on a space $(X, \mathcal{B})$, if $\mu = \mu_1 - \mu_2$ then $\mu_2 = 0$.