Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
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: Yes, that's exactly right. I apologize for being somewhat too terse. – t.b. Jan 16 '11 at 20:20
@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
@Matt: exactly! – t.b. Nov 22 '11 at 22:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.