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Let $(a_n)_{n=1}^{\infty}$ be an enumeration of $\mathbb{Q} \cap [0,1]$. Define $B_n := \left\{x \in [0,1] : x \in \left( a_n \pm \frac{1}{2 \cdot 3^n}\right)\right\}$, $C := \bigcup_{n=1}^{\infty}B_n$.

Since $\mu(B_1) \geq 1/6$ and $\sum_{n=1}^{\infty}\mu(B_n) \leq 1/2$, the indicator $\chi_C$ integrates to some value between $1/6$ and $1/2$ using the Lebesgue integral. Since each set in a partition of $[0,1]$ will contain some values $x$ with $\chi_C(x) = 1$, all upper Darboux sums of $\chi_C$ are 1, and so $\chi_C$ isn't Riemann integrable.

My question: Does there exist some $f : [0,1] \to \mathbb{R}$ with $f = \chi_C$ almost everywhere such that $f$ is Riemann integrable?

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I believe the answer is no, and my thinking is along these lines:

I assume you're familiar with the result that Riemann-integrable functions are precisely those that are continuous almost-everywhere. So take $E \subseteq [0, 1]$ to be the set of points where $f$ is continuous.

Since $f = \chi_C$ a.e., we can say that $f = \chi_C$ on a set $F$ of full measure. Then the restriction of $\chi_C$, considered as a function on the set $E \cap F$, is continuous.

It follows that no point in the set $C^c \cap E \cap F$ can be the limit of points from $C \cap E \cap F$, or else continuity would fail. (Because $0 < \lambda(C) < 1$, and $E \cap F$ has full measure, both $C^c \cap E \cap F$ and $C \cap E \cap F$ are nonempty.)

However, note that $C$ is a dense open set in $[0, 1]$, and you can verify that the intersection of a dense open set and a set of full-measure is still dense. So $\overline{C \cap E \cap F} = [0, 1]$, the aforementioned property fails, and it seems that no such $f$ exists.

(I typed this up in a rush and late at night; please let me know if you think there are any mistakes.)

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  • $\begingroup$ Excellent. If I understood you correctly, the argument is that $C \cap E \cap F$ contains all its limit points, i.e., is its own closure, contradicting the fact that the closure of $C \cap E \cap F = [0,1]$. $\endgroup$
    – jvh_ch
    Commented Nov 8, 2023 at 11:41

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