Reimann Integral definition confusion.

  • Thread starter sid9221
  • Start date
  • #1
sid9221
111
0
My lecture notes say:

Let f:[a,b]->R be bounded.
F is said to rienmann integrable if:

[tex] L=\int_{a}^{b} f(x)=U [/tex]

where :

L=Sup(L(f,P))

and

U=Inf(U,(f,P))


but everywhere else(internet) there's a definition with epsilon.
I have the epsilon stuff later under "riemann criterion" so was wondering if the above definition is okay or I copied something wrong
 

Answers and Replies

  • #3
sid9221
111
0
Never heard of a darboux integral will this definition suffice if asked for a riemann integral?
 
  • #4
Bacle2
Science Advisor
1,089
10
My lecture notes say:

Let f:[a,b]->R be bounded.
F is said to rienmann integrable if:

[tex] L=\int_{a}^{b} f(x)=U [/tex]

where :

L=Sup(L(f,P))

and

U=Inf(U,(f,P))


but everywhere else(internet) there's a definition with epsilon.
I have the epsilon stuff later under "riemann criterion" so was wondering if the above definition is okay or I copied something wrong

Actually, this is an equality in the limit, i.e., you need to be able to make the difference

L-U as small as possible, i.e., show that there are partitions such that, for all e>0,

L-U<e .

Or, if you really want to go over the top, use net-convergence on the partitions.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
43,021
970
My lecture notes say:

Let f:[a,b]->R be bounded.
F is said to rienmann integrable if:

[tex] L=\int_{a}^{b} f(x)=U [/tex]

where :

L=Sup(L(f,P))

and

U=Inf(U,(f,P))


but everywhere else(internet) there's a definition with epsilon.
I have the epsilon stuff later under "riemann criterion" so was wondering if the above definition is okay or I copied something wrong
Yes, the "epsilon stuff" is hidden in the sup and inf- they are limits. If they are equal, then f is "integrable" and their common value is the integral of f.

Here is an example in which it does NOT work and so f is not integrable: let f(x)= 0 if x is rational and 1 if x is irrational. Now, divide the interval from a to b into n sub-intervals. Inside any such sub-interval, there exist both rational and irrational numbers so f(x) has values of both 0 and 1 inside each interval. That means that each [itex]L(f, P)[/itex], the sum of the minimum height and base of each interval, is 0 and each [itex]U(f, P)[/itex], the sum of the maximum height and base of each interval is 1. The "sup" is 1 and the "inf" is 0. They are not the same so f is not integrable.
 
  • #6
Karamata
60
0
In real analysis, a branch of mathematics, the Darboux integral or Darboux sum is one possible definition of the integral of a function. Darboux integrals are equivalent to Riemann integrals, meaning that a function is Darboux-integrable if and only if it is Riemann-integrable, and the values of the two integrals, if they exist, are equal.

Darboux sum is important here.

We define upper Darboux sum and lower Darboux sum, and say that function is Riemann-integrable iff [itex]\displaystyle\sup_{P}L_{f,P} = \displaystyle\inf_{P}U_{f,P}[/itex] iff [itex]\forall \varepsilon>0 ~\forall P ~ \exists \delta>0~ \mbox{if} ~\lambda(P)<\delta ~\mbox{then} ~|U_{f,P}-L_{f,P}|<\varepsilon[/itex], where [itex]\lambda =
\displaystyle\max_{1\le k \le n} |x_{k}-x_{k-1}|[/itex]

We supposed that [itex]f[/itex] is bounded.
 

Suggested for: Reimann Integral definition confusion.

Replies
6
Views
209
Replies
1
Views
182
Replies
5
Views
138
Replies
2
Views
311
Replies
18
Views
929
Replies
13
Views
657
Replies
23
Views
365
  • Last Post
Replies
7
Views
136
Replies
4
Views
338
Top