# Can the Lebesgue integral be defined so that it can be applied to non-Lebesgue measurable subsets of $\mathbb{R}^d$?

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?

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 $$0$$.

• Could there be a non-trivial function $f$ for which $f \cdot \chi_V$ is measurable?
– user77614
Feb 6, 2021 at 1:15
• @user1975053 You could pick a measurable set $E$ containing $V$ and set $f : = \chi_{E \setminus V}$. Feb 6, 2021 at 1:39
• But would $E \setminus V$ be measurable?
– user77614
Feb 6, 2021 at 9:06
• @GEdgar How could linearity fail? Wouldn’t always be $\chi_{M \cup N}=\chi_M + \chi_N$?
– user77614
Feb 6, 2021 at 9:11
• 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. Feb 6, 2021 at 12:22