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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$f$?

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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.

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