We can define the integral of a function $f$ on any $M \subseteq \mathbb{R}^d$ as $\int_{\mathbb{R}^d} f \cdot \chi_M$, irrespective of whether $M$ is Lebesgue-measurable or not. It works fine for the zero function on the Vitali set $V$, so that $\int_V 0=0$. However, if we follow the measure theory approach, $\int_V 0$ is simply not defined. Is this not a shortcoming of the measure theory approach to the Lebesgue integral? Could there not be some non-trivial function $f$ with $f \cdot \chi_M$ integrable over $\mathbb{R}^d$, even if $M$ is non-measurable?


1 Answer 1


Of course you can use an "upper integral" or a "lower integral". But if you do this, then linearity can fail. That is, the property $\int_M f + \int_N f = \int_{M \cup N} f$ for disjoint sets $M,N$ could fail.

Your case of the $0$ function on the Vitali set $V$, though, is not a problem. We define $\int_V 0$ to be $\int_{\mathbb R} 0\cdot \chi_V$. But since $0 \cdot \chi_V$ is $0$ on all of $\mathbb R$, this integral is computed as the integral of the measurable function $0$.

  • $\begingroup$ Could there be a non-trivial function $f$ for which $f \cdot \chi_V$ is measurable? $\endgroup$
    – user77614
    Feb 6, 2021 at 1:15
  • $\begingroup$ @user1975053 You could pick a measurable set $E$ containing $V$ and set $f : = \chi_{E \setminus V}$. $\endgroup$
    – Oiler
    Feb 6, 2021 at 1:39
  • $\begingroup$ But would $E \setminus V$ be measurable? $\endgroup$
    – user77614
    Feb 6, 2021 at 9:06
  • $\begingroup$ @GEdgar How could linearity fail? Wouldn’t always be $\chi_{M \cup N}=\chi_M + \chi_N$? $\endgroup$
    – user77614
    Feb 6, 2021 at 9:11
  • 1
    $\begingroup$ Yes $\chi_{M \cup N}=\chi_M + \chi_N$, but linearity of the integral is proved for measurable functions only. This is why the upper integral and lower integral (defined even for non-measurable functions) are not very useful. $\endgroup$
    – GEdgar
    Feb 6, 2021 at 12:22

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