Take the 2-minute tour ×
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.

Let $(X, \mathcal{S},\mu)$ be a measure space, and let $f: X \to \mathbb{C}$ be a function. Then $f$ is integrable if Re$f$ and Im$f$ are integrable and $\int f := \int$Re$f+i\int$Im$f$.

It is easy to show that:

Re$f$ and Im$f$ are measurable and $|f|$ is integrable $\Rightarrow$ $f$ is integrable.

But why can't we omit the measurability of Re$f$ and Im$f$?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

$|f|$ being integrable (or measurable) alone cannot give information about whether $Re f$ or $Im f$ being measurable. For example, take $A$ to be your favorable nonmeasurable set, $Re f$ to be the characteristic function of $A$, and $Im f$ to be the characteristic function of the complement of $A$. Then $|f|$ is really the constant function 1, which is integrable if the total space has finite measure.

share|improve this answer

Your Answer

 
discard

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.