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It is known that if $f$ is Riemann integrable function then it is Lebesgue integrable, in particular $\ f:\left(\mathbb{R},\mathcal{L}\left ( \mathbb{R} \right),\lambda \right) \longrightarrow \mathbb{R} \ \ $ is measurable. In the proof of this theorem the completeness of $\mathcal{L} \left ( \mathbb{R} \right)$ is important.

My question: Is there any Riemann intagrable function $f:\left(\mathbb{R},\mathcal{B}\left ( \mathbb{R} \right), \lambda \right) \longrightarrow \mathbb{R} \ \ $ wiche is not $\mathcal{B}\left ( \mathbb{R} \right)$-measurable ?

Note that $\mathcal{B}\left ( \mathbb{R} \right)$ is not complet with respect to $\lambda$.

G. Chiusole
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Student
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1 Answers1

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Let $C$ be the Cantor set, and $D \subseteq C$ such that $D$ is NOT Borel measurable.

(The construction of such $D$ can be easily found. For instance: Lebesgue measurable set that is not a Borel measurable set).

Now consider $\chi_D$ the indicator function of $D$, that is $\chi_D(x)=1$ if $x\in D$ and $\chi_D(x)=0$ if $x\in \mathbb{R} \setminus D$.

It is easy to see that $\chi_D$ is not a Borel measurable function, because $\chi_D^{-1}(\{1\}) = D$.

On the other hand, $\chi_D$ is continuous in the open set $\mathbb{R} \setminus C$. So the points of discontinuities of $\chi_D$ is contained in $C$ and we know that $\lambda(C)=0$. So $\chi_D$ is Riemann integrable.

Ramiro
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