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There are many ways to define the Riemann Integral. I am using this one, where I denote $\sigma(f,P^{*})$ the Riemann Sum relative to a tagged partition $P^{*}$:

$\textbf{Definition}$

We say that a function $f:[a,b] \to \mathbb{R}$ is Riemann-Integrable if exist the limit:

$I=\lim_{||P|| \to 0} \sigma(f,P^{*})$

and then we write $I=\int_{a}^{b}f(x)dx$.

The limit exist in the sense that given $\epsilon > 0$, there is $\delta > 0$ such that for any partition $P$ of $[a,b]$ with $||P|| < \delta$ and for any tagged partition $P^{*}$, we have:

$$|\sigma(f,P^{*}) - I| < \epsilon$$

By definition, if $f$ is integrable in $[a.b]$, then given $\epsilon < 0 $ exists two tagged partition $P^{*}$ and $P^{**}$ such that:

$$|\sigma(f,P^{*}) - I| < \epsilon / 2$$

$$|\sigma(f,P^{**}) - I| < \epsilon / 2$$

hence, $|\sigma(f,P^{*}) - \sigma(f,P^{**})| < \epsilon$. Therefore, is a necessary condition for integrability that one can find a partition $P$ such that two Riemann Sums relative to $P$ are very close together, no matter what the scalars that we pick in $P^{*}$ and $P^{**}$.

I am pretty sure that this is a sufficient condition as well. But I have any ideas how to prove it. Can anyone help with this?

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  • $\begingroup$ It seems valid since one can choose a Riemann sum arbitrarily close to a Darboux sum. $\endgroup$ – Tony Piccolo Nov 24 '18 at 6:58
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There is such a sufficient (Cauchy) condition. The correct statement is:

Suppose for any $\epsilon > 0$ there exists $\delta > 0$ such that $|\sigma(f,P) - \sigma(f,P')| < \epsilon$ for all partitions $P$ and $P'$ with $\|P\|, \, \|P'\| < \delta$ and for any choice of tags. Then $f$ is Riemann integrable.

In proving this one must first show the existence of a viable candidate for the value of the integral and then show that it satisfies the required definition.

First -- and I will leave this to you for now -- construct a decreasing sequence of positive numbers $\delta_n$ and partitions $P_n$ with $\|P_n\| < \delta_n$ such that for any partition $P$ with $\|P\| < \delta_n$ we have (for any choice of tags)

$$|\sigma(f,P) - \sigma(f,P_n)| < 1/n$$

Hence, if $m \geqslant n$ then $|\sigma(f,P_m) - \sigma(f,P_n)| < 1/n$. The sequence $\sigma(f,P_n)$ is a Cauchy sequence and must converge to a real number $I$.

To show that $I$ satisfies the definition of the integral $\int_a^b f(x) \, dx$, for any $\epsilon >0$ take $n$ such that $1/n < \epsilon/2$ and sufficiently large such that $|\sigma(f,P_n) - I| < \epsilon/2$. If $P$ is a partition with $\|P\| < \delta_n$, then it follows that

$$|\sigma(f,P) - I| \leqslant |\sigma(f,P) - \sigma(f,P_n)| + |\sigma(f,P_n) - I| < 1/n + \epsilon/2 < \epsilon $$

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It is a modified version of the standard Cauchy criterion.

The crucial statement is

Let $f:[a,b] \to \mathbb{R}$ be a bounded function.
If for every $\varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $|\sigma (f,P^*)- \sigma (f,P^{**})|<\varepsilon$ for any Riemann sums $\sigma (f,P^*)$ and $\sigma (f,P^{**})$ for $f$ associated with $P$, then $S(f,P)-s(f,P)<3\varepsilon$.

It is about using the fact that $S(f,P)$ and $s(f,P)$ are respectively the supremum and the infimum of the set of Riemann sums for $f$ associated with $P$. The triangular inequality does the rest.

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