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Can we approach the Riemann integral with measure theory?

That is: can we find a measure $\mu$ defined on a $\sigma$-algebra of $\mathbb{R}$ such that a function is $\mu$-integrable if and only if it is Riemann integrable, and that the integral $\int f d\mu$ is equal to the corresponding Riemann integral. If so can we extend this to improper Riemann integrals? What about Riemann integration in $\mathbb{R}^n$?

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up vote 8 down vote accepted

No.

Singletons would have to be measurable in this $\sigma$-algebra since their characteristic functions are Riemann integrable, alternatively because $\{x\} = \bigcap_{n=1}^\infty \left[x-\frac{1}{n},x+\frac{1}{n}\right]$ shows that they are a countable intersection of measurable sets. Therefore the characteristic function of every countable set would have to be Riemann integrable, but $\chi_{[0,1] \cap \mathbb{Q}}$ provides a counterexample.

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