Let $f= f^+-f^-$. Show that $f$ is Borel measurable if and only if $f^+, f^-$ are both Borel measurable.
Attempt:
Suppose that $f^+$ and $f^-$ are both Borel measurable. Then their difference $f^+-f^-$ is also Borel measurable because the difference of two Borel measurable functions $a(x),b(x)$ is Borel measurable because $\{a(x)>\alpha+b(x)\}$ can be written as a union of Borel measurable sets.
For the other direction, I'm not sure where to start. I thought of trying to rewrite $f=f^+-f^-$ as the difference of two Borel measurable functions again, but of course we don't know that either $f^+, f^-$ are Borel measurable. A hint to get started would be welcomed. Thanks.
(By the way, $f^+, f^-$ is notation used in lattices, where $f^-=-(f\wedge 0)$ and $f^+=f \vee 0$)
Let $U_a = (a, \infty)$. We note that $$ (f^+)^{-1}(U_a) = \\ \{f^+(x) > a\} = \\ \{\max\{f(x),0\} > a \} $$ this gives us $f^{-1}(U_a)$ when $a \geq 0$ and $\Bbb R$ otherwise. In either case, we conclude $(f^+)^{-1}(U_a)$ is Borel.
The proof for $f^-$ is similar.