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Jul 7, 2022 at 11:25 comment added Gerald Edgar @user7427029 an explicit example, including constructing a fat Cantor set, is in the paper mentioned by Quaochu Yuan ... jstor.org/stable/2589023
Jul 7, 2022 at 9:41 comment added user7427029 Could you specify a concrete $g$ as proof of its existence? @X.M. Du: A nice question for your freshman could be: "Why does the above counterexample, especially $g$, not contradict the Riemann integrability of (most of the) continuous functions described here?": math.stackexchange.com/questions/2580703/…
Jun 21, 2016 at 5:10 comment added Paramanand Singh If I compare your answer with Qiaochu Yuan's answer I see that you have a counterexample with continuous and increasing $g$ whereas Yuan's answer says that if $g$ is monotone the result is true. I want to know if the result is true when $g$ is monotone. Also if possible answer on this question math.stackexchange.com/q/1833028/72031
Jan 31, 2013 at 3:28 comment added user20948 $C$ could be just ${0}$ as the article Qiaochu Yuan mentioned does. We, our freshmen, could understand a rather strong criterion of Riemann integrability more elementary than Lebesgue's characterization, which is out of ability of us, at least, most of us: Suppose $f$ is bounded on $[a,b]$, $f$ is Riemann integrable if and only if $\forall\epsilon,\eta>0$, there's some partition $P=\{a=x_0,\dotsc,x_n=b\}$ of $[a,b]$ such that $\sum_{M_k-m_k>\eta}\delta x_k<\epsilon$, where $M_k=\sup_{x_{k-1}\le x\le x_k}f(x),m_k=\inf_{x_{k-1}\le x\le x_k}f(x),\delta x_k=x_k-x_{k-1}$.
Apr 2, 2010 at 8:41 vote accept X.M. Du
Apr 1, 2010 at 12:09 history answered Gerald Edgar CC BY-SA 2.5