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I am currently self-studying real analysis using Rudin's book. Rudin defines Riemann integrability as follows:

Let $f: [a,b]\rightarrow \mathbb{R}$, and let $P$ be a partition of $[a,b]$. Then $f$ is Riemann integrable on $[a,b]$ if $\inf U(P,f) = \sup L(P,f)$, where $U(P,f)$ and $L(P,f)$ denote the upper and lower Darboux sums of $f$ (with respect to the partition P), respectively, and the $\sup$ and $\inf$ are taken over all partitions $P$.

The following is a statement that I assumed to be true when I took calculus, but now I am not so sure:

Let $f$ be Riemann integrable on $[a,b]$. Given a partition $P$ of $[a,b]$, let $\left\| P \right\|$ denote the norm of $P$ (i.e. the length of the longest segment in $P$). Is it then true that as $\left\| P \right\| \rightarrow 0$, $U(P,f)-L(P,f)\rightarrow 0$? That is, is it true that for any $\epsilon>0$, there exists $\delta>0$ such that $\left\| P \right\|<\delta \Rightarrow U(P,f)-L(P,f)<\epsilon$?

It is easy to show that this result holds when $f$ is continuous on $[a,b]$, but I'm not sure if it holds for all Riemann integrable functions.

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The Riemann integral defined as $I = \sup_P L(P,f) = \inf_P U(P,f)$ is equivalent to the integral defined as $I = \lim_{\|P\| \to 0} S(P,f)$ where $S(P,f)$ is a sum of the form $\sum_{j=1}^n f(c_j)(x_j - x_{j-1})$. When $f$ is Riemann integrable, that limit is unique regardless of how the tags $c_j \in [x_{j-1},x_j]$ are chosen or how the partition points are distributed as $\|P\| = \max_{1 \leqslant j \leqslant n} (x_j - x_{j-1}) \to 0$. This is proved here.

As a consequence we also have $U(P,f)-L(P,f) \to 0$ as $\|P\| \to 0$. This follows because for any $\epsilon > 0$ and any partition interval $I_j =[x_{j-1},x_j]$ we can select tags such that

$$\sup_{I_j} f(x) - \epsilon/(2(b-a)) < f(\xi_j) \leqslant \sup_{I_j} f(x), \\\inf_{I_j} f(x) \leqslant f(\eta_j) < \inf_{I_j} f(x) +\epsilon/(2(b-a)), $$

and, forming Darboux and Riemann sums,

$$U(P,f) - L(P,f) < S(P,f,\{\xi\})-S(P,f,\{\eta\}) + \epsilon$$

The RHS converges to $\epsilon$ as the partition norm $\|P\| \to 0$, since both Riemann sums approach the integral. Hence,

$$0 \leqslant \lim_{\|P\| \to 0}[U(P,f) - L(P,f)] \leqslant \epsilon.$$

Since $\epsilon$ is an arbitrary positive real number, the result follows.

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  • $\begingroup$ It's not a particularly easy thing to prove and very few books do, but it is true. $\endgroup$
    – RRL
    Commented Jun 11, 2018 at 2:38

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