Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Following was asked in a question by jpv (which is in turn pointed out by t.b. to me):

Let $(X, \mathcal{F}, \mu)$ be a measure space and $(Y,d)$ be a separable metric space ($d$ is the metric). If $f:(X,\mathcal{F}) \rightarrow (Y, d)$ is a $\mu$-measurable function prove that there exists an $\mathcal{F}$ measurable function which coincides with $f$ everywhere except on a $\mu$-negligible set.

EDIT: The textbook is "Functions of Bounded Variation and Free Discontinuity Problems" by Luigi Ambrosio et. al.

I was wondering what is the definition of a $\mu$-measurable function,and how it is different from $\mathcal{F}$ measurable function?

I looked it up in the book mentioned, but cannot find where it is.

Is a $\mu$-measurable function defined as a function which is measurable when it is restricted to the complement of a measure zero measurable subset of $X$?

Thanks and regards!

share|cite|improve this question
up vote 1 down vote accepted

On p. 6 of that textbook, it defines a $\mu$-measurable function as one which is measurable on the unique sigma algebra associated with the completion of the measure $\mu$.

As an aside, in Serge Lang's book Real and Functional Analysis, he defines a $\mu$-measurable function $f$ as one which is the pointwise limit of a sequence of simple functions almost everywhere. Not being an expert on abstract measure theory myself, I would be interested in seeing if these two conditions are equivalent or related. This is in the context of Bochner integration, so at the very least we'd first have to restrict ourselves just to real-valued $f$.

share|cite|improve this answer
They are the same for countably additive measures. See Dunford & Schwartz 1958, who develops measurable functions starting from finitely additive measures. – Rabee Tourky Feb 5 '13 at 22:38
Thanks, this is great to know. I've always been confused about the different definitions and conventions in measure theory. – Christopher A. Wong Feb 6 '13 at 0:05

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.