Let $f:[a,b] \to \mathbb{R}$ be Riemann integrable, where we use the usual definition using the upper and lower Darboux sums. For a given partition $P$ defined by $x_0 = a < x_1 < \ldots < x_n = b$ we denote the upper sum by $U(f,P)$ and the gap $|P| = \sup_{1 \le i \le n}(x_i - x_{i-1})$. It is easy to show that if $P_n$ is a sequence of partitions with $|P_n| \to 0$ and $f$ is continuous then we must have $$\int_a^b f = \lim_n U(f,P_n).$$ Does this same statement hold true if $f$ is discontinuous but still integrable?
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$\begingroup$ Yes. Hint: Points of discontinuity of a Riemann integrable function is a measure zero set, and a measure zero set can be put inside an open set of sufficiently small length. $\endgroup$– SurajitMay 1, 2019 at 18:46
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1$\begingroup$ @uniquesolution No. Enumerate the rationals by $\{x_n\}_{n\in\mathbb{N}}$, and for an arbitrarily fixed $\epsilon >0$, consider the open neighbourhood $\bigcup_{n\in\mathbb{N}}\Big(x_n-\frac{\epsilon}{2^{n+1}},x_n+\frac{\epsilon}{2^{n+1}}\Big)$ of rationals. Length of this open set is $\sum_n\frac{2\epsilon}{2^{n+1}}=\epsilon$. $\endgroup$– SurajitMay 1, 2019 at 19:15
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$\begingroup$ @Surajit - you are right, my bad. $\endgroup$– uniquesolutionMay 1, 2019 at 19:17
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$\begingroup$ @uniquesolution $(0,1)$ is the only open interval containing $\mathbb{Q} \cap (0,1)$, but not every open set is an interval. $\endgroup$– ChocolateAndCheeseMay 1, 2019 at 19:17
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$\begingroup$ @ChocolateAndCheese - yes, that is correct. $\endgroup$– uniquesolutionMay 1, 2019 at 19:19
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1 Answer
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The answer is yes because $$\lim_n U(f,P_n)=U(f)$$ anyway if $|P_n|\rightarrow 0\,.$
In order to prove the statement one can use the standard inequality $$U(f,P) \le U(f,Q) + 2Mq|P|$$ where $P$ and $Q$ are partitions of $[a,b]$, $M$ is an upper bound of $|f|$ and $q$ is the number of partition points of $Q\,.$
Then, if $\varepsilon>0$, it is easy to show that $$U(f,P_n) < U(f) + \varepsilon$$ if $n$ is enough large.
answered May 8, 2019 at 15:54