A measurable function (with complete measure) is sum of two other functions Let $(\Omega, \mathcal A, \mu)$ be a measure space and let $\overline{\mu}$ denote the completion of $\mu$. I have to show that if $f \colon \Omega \to \mathbb R$ is $\overline{\mu}$-measurable then $f = f_1 +f_2$, where $f_1$ is $\mu$-measurable and $f_2 = 0$ ($\overline{\mu}$-)almost everywhere.
I have no idea how to start. It seems that $f_2$ has to be nonzero somewhere (because we have completed the measure), but I'm not sure where.

Hint:


*

*Prove the claim for indicator functions, i.e. $f=1_A$ with $A \in \bar{\mathcal{A}}$.

*Extend it to simple functions.

*Let $f \geq 0$ be $\bar{\mu}$-measurable. Then there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions which are $\bar{\mathcal{A}}$-measurable and satisfy $f_n \to f$ as $n \to \infty$. Use step 2 to conclude that the claim holds.

*For general $f$ write $f=f^+ - f^-$ and apply step 3 to $f^{\pm}$.

