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I have some trouble with the Riemann integral, specifically, the definition of it in an article on wikipedia.

We say that the Riemann integral of $f$ equals s if the following condition holds:

For a given $\varepsilon>0$, there exists $\delta$ such that for any tagged partition $x_0,\cdots,x_n$ and $t_0,\cdots,t_{n-1}$ whose mesh is less than $\delta$, we have $$\left|\sum_{i=0}^{n-1}f(t_i)(x_{i+1}-x_i)-s\right|<\varepsilon.$$

Unfortunately, this definition is very difficult to use. It would help to develop an equivalent definition of the Riemann integral which is easier to work with. We develop this definition now, with a proof of equivalence following.(?) Our new definition says that the Riemann integral of f equals s if the following condition holds:

For all $\varepsilon>0$, there exists a tagged partition $x_0,\cdots,x_n$ and $t_0,\cdots,t_{n-1}$ such that for any refinement $y_0,\cdots,y_m$ and $s_0,\cdots,s_{m-1}$ of $x_0,\cdots,x_n$ and $t_0,\cdots,t_{n-1}$, we have $$\left|\sum_{i=0}^{m-1}f(s_i)(y_{i+1}-y_i)-s\right|<\varepsilon.$$

  • It is easy to show that the first definition implies the second.
  • To show that the second definition implies the first, it is easiest to use the Darboux integral. First one shows that the second definition is equivalent to the definition of the Darboux integral; for this I want you to give me some hints. However, I know that it is also easy to show that a Darboux integrable function satisfies the first definition.

The original article says that the second definition is equivalent to the definition of the Darboux integral; for this see the article on Darboux integration. Well, I couldn't find anything useful information.

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  • $\begingroup$ Typically the number of division points in a "refinement" of a partition is (or at least may be) greater than the number in the partition being "refined". However above you have the same number $n$ of points for each. That doesn't seem like a refinement in the usual sense. Was the above a direct quote from a source, or something you re trying to come up with? $\endgroup$ – coffeemath Aug 2 '13 at 9:35
  • $\begingroup$ @coffeemath I edited the ques. $\endgroup$ – Brooks Aug 2 '13 at 9:37
  • $\begingroup$ @coffeemath Thanks. But I didn't get you idea. Could you explain it again to me? $\endgroup$ – Brooks Aug 2 '13 at 10:12
  • $\begingroup$ I just looked at the article you link to in your question. (I think you should mark off the part of your question which is quoted from that source separately). Anyway on that link is a full proof of the equivalence of the two definitions, so I don't see much advantage of reproducing it here; maybe your question should focus on where in their proof you can't follow the argument. $\endgroup$ – coffeemath Aug 2 '13 at 10:25
  • $\begingroup$ I edited the ques. $\endgroup$ – Brooks Aug 2 '13 at 10:46
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Your definition is

$(1)$ $f\in \mathscr R[a,b]$ with integral $\int_a^b f$ if and only if for each $\epsilon >0$ there exists $\delta >0$ such that for any partition $P$ with $\lVert P\rVert <\delta$ we have that $$\left|\int_a^b f-\sum(f,P)\right|<\epsilon$$

where $\sum(f,P)$ means the Riemann sum of $f$ with respect to $P$. The alternative definition is

$(2)$ $f\in \mathscr R[a,b]$ with integral $\int_a^b f$ if and only if for each $\epsilon >0$ there exists a partition $P_\epsilon$ such that for any partition $P$ a refinement of $P_\epsilon$ we have that $$\left|\int_a^b f-\sum(f,P)\right|<\epsilon$$

We would like to show one implies the other. First

$(1) \implies (2)$ Suppose $(1)$ holds. Since refinements can only decrease the mesh, and $\delta$ depends on $\epsilon$, the claim follows: for each $\epsilon>0$ take a partition $P_\epsilon$ with mesh $\delta'<\delta$ given by the above. Then any refiniment will have mesh at most $\delta'$ which will be less than $\delta$, and hence $(2)$ will hold.

$(2)\implies (1)$ This one is the tricky one. Here you can find a proof


An easy characterization of Riemann integrability is the following, which doesn't requiere that we know what the value of integral is:

$f:[a,b]\to\Bbb R$ is Riemann integrable over $[a,b]$ if and only if for each $\epsilon >0$ there exist a partition $P_\epsilon$ such that for any refinement $P$ of $P_\epsilon$ $$U(f,P)-L(f,P)<\epsilon$$

Moreover, this is equivalent to $$\overline{\int_a^b} f=\underline{\int_a^b}f$$

so if you're able to prove the above and evaluate any upper or lower integral, you're done.

In fact, this applies to the Riemann Stieljes integral whenever the integrator $\alpha$ is montone.

Let $\alpha:[a,b]\to\Bbb R$ be monotone. Let $f:[a,b]\to\Bbb R$. Then the following are equivalent:

$(1)$ The function $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ over $[a,b]$

$(2)$ For each $\epsilon >0$ there exists a partition $P_\epsilon$ such that for any refinement $P$ of $P_\epsilon$ $$U(f,\alpha,P)-L(f,\alpha,P)<\epsilon$$

$(3)$ $$\overline{\int_a^b} f=\underline{\int_a^b}f$$

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  • $\begingroup$ thanks, I think of your answer over and do it myself. however, it is solved owe to yoy. $\endgroup$ – Brooks Dec 31 '15 at 5:18

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