"Let μ 0 ( A ) = μ ( A ∪ A 0 ) ,...

"Let ${\mu }_0(A)=\mu (A\cup A_0),A\in \mathcal{F}$ be a measure. Show that if
$\int fd\mu$
exists, then
${\int }_{A_0}fd\mu =\int f{d}_{{\mu }_0}$"
However, I think there's a typo in this exercise. ${\mu }_0$ isn't even a measure when $\mu (A_0)\ne 0$, since ${\mu }_0(\varnothing )=\mu (A_0)$. If it were ${\mu }_0(A)=\mu (A\cap A_0)$, which I think that's what the author meant, I could prove it in the following way:
For simple functions,
${\int }_{A_0}fd\mu =\int f{\mathbf{1}}_{A_0}d\mu =\sum _{i=1}^n{f}_i{\mathbf{1}}_{A_0}\mu (A_i)=\sum _{i=1}^n{f}_i\mu (A_i\cap A_0)=\sum _{i=1}^n{f}_i{\mu }_0(A_i)=\int fd{\mu }_0$
For a non-negative measurable function $f$, there's a non-decreasing sequence of simple functions $(s_n{)}_{n\in \mathbb{N}}$ such that $s_n\to f$, then
${\int }_{A_0}fd\mu =\int f{\mathbf{1}}_{A_0}d\mu =\underset{n\to \mathrm{\infty }}{lim}\int s_n{\mathbf{1}}_{A_0}d\mu =\underset{n\to \mathrm{\infty }}{lim}\int s_{n}d{\mu }_0=\int fd{\mu }_0$
And finally, for any measurable function, we just use $f=f^{+}-f^{-}$. So, which one is it, $\mu (A\cup A_0)$ or $\mu (A\cap A_0)$?