# convergence of Riemann integral as partition size goes to zero

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?

• 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. May 1, 2019 at 18:46
• @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 $\sum_n\frac{2\epsilon}{2^{n+1}}=\epsilon$. May 1, 2019 at 19:15
• @Surajit - you are right, my bad. May 1, 2019 at 19:17
• @uniquesolution $(0,1)$ is the only open interval containing $\mathbb{Q} \cap (0,1)$, but not every open set is an interval. May 1, 2019 at 19:17
• @ChocolateAndCheese - yes, that is correct. May 1, 2019 at 19:19

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.