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  "body": "\n## Counterexample\n\nLet $f \\colon [0, 1] \\to \\mathbb{R}$ be the [[Dirichlet function|0T4EJ8ZJM9]]: $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$.\nEach subinterval of positive length contains both rational and irrational numbers. So for each partition $P$, $U(f, P) = 1$ and $L(f, P) = 0$.\nThe upper [[Darboux integral|09FQT057T7]] is $1$ and the lower one is $0$. So $f$ is [[not Riemann integrable|021XJAQVDC]].\n\n## Remarks\n\nThe function is bounded, so the failure is not caused by unboundedness.\nThe function is Lebesgue integrable, with Lebesgue integral $0$, since $\\mathbb{Q}$ has measure zero.\nIt is the pointwise limit of Riemann integrable functions. So pointwise limits do not keep Riemann integrability.\n\n## Explanations\n\n### L1 human\n\nEvery Riemann sum of this function is $1$ or $0$ or in between, depending only on the choice of tags.\nSo the sums cannot settle down to one number.\n",
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  "msc": [
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  "name": "The Dirichlet function is not Riemann integrable",
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