Change the values of a measurable function on a negligible set If $f$ is a $\mu$-measurable function and I change its values on a $\mu$-negligible set, i.e. on $Y \subset Z$ with $\mu (Z) = 0$, why is $f$ still measurable?
 A: This is not true if the measure is not complete. Simply take a non-measurable subset $Y$ of a measurable null-set $Z$ and look at the characteristic function of $Y$, which is clearly not measurable but agrees with the (measurable) zero function outside of $Y$.
However, your modified $f$ will be measurable with respect to the completion $\tilde{\mu}$ of the measure and agree with $f$ outside of a $\tilde{\mu}$-null set, so the $\tilde{\mu}$-integral will remain unchanged by such a modification and that's enough for most purposes.
To make this more precise, let $g$ be the modified function and $Y = \{x\,:f(x) \neq g(x)\}$. By hypothesis, $Y \subset Z$ with $Z$ measurable and $\mu(Z) = 0$. Observe that $g = [X \smallsetminus Y] \cdot f + [Y] \cdot (g - f)$. With respect to the completion $Y$ and $X \smallsetminus Y$ are measurable because $\tilde{\mu}(Y) = 0$. Note that on a complete space every function which is nonzero only inside a null-set is measurable. Therefore $g$ is the sum of two $\tilde{\mu}$-measurable functions and hence measurable.
