# 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?

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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.
Theo, many thanks, it took me a while to understand. In other words: if $f$ is the constant function $f(x) = 0$ and $Z$ is a non-measurable, negligible set then changing $f$ to $1$ on $Z$ makes it a non-measurable function. On the other hand, if I make my space complete by defining the standard extension (all unions $\hat{Y} = Y \cup Y_0$, $Y$ measurable, $Y_0$ negligible and $\mu^\prime (\hat{Y}) := \mu(Y)$) then there are no non-measurable, negligible sets anymore, so changing any $f$ on a negligible set leaves it measurable. –  Rudy the Reindeer Jan 16 '11 at 20:05
@Matt: To connect all this back to your previous question about $L^{1}$-spaces it is an illuminating exercise to prove that the obvious map $L^{1}(\mu) \to L^{1}(\tilde{\mu})$ is an isometric isomorphism, i.e., it is well-defined, linear, onto and $\|f\|_{L^{1}(\mu)} = \|f\|_{L^{1}(\tilde{\mu})}$ for all $f \in L^{1}{(\mu)}$. –  t.b. Jan 16 '11 at 20:28
Taking the obvious map $\varphi: f \mapsto f$, linear and surjective are "obvious". It's well-defined: Let $f \neq f^\prime$ on $U$ with $\mu(U) = 0$. Then $\tilde{\mu}(U) = 0$ and so $f = f^\prime$ $\tilde{\mu}$-a-e. Finally, $$\| f \|_{L^1 (\mu)} = \int_X |f| d \mu = \sup \{ \int_X s d \mu \mid s \text{ step function }, s \leq |f| \} =$$ $$\sup \{ \sum_{i=1}^n \alpha_i \mu(Y_i) \mid \text{ for some } \alpha_i \} = \sup \{ \sum_{i=1}^n \alpha_i \tilde{\mu} \mid \text{ for some } \alpha_i \} = \| f \|_{L^1(\tilde{\mu})}$$. –  Rudy the Reindeer Nov 22 '11 at 22:20