# Reimann Integral definition confusion.

• sid9221
The upper Darboux sum is the supremum of the length of all interval that contain x within the domain, and the lower Darboux sum is the infimum of the length of all interval that contain x within the domain. We also say that if x is in an interval, then the Darboux sum of that interval is greater than or equal to the lower Darboux sum of any other interval that contains x. If we try to find the Riemann integral of f, we need to make sure that the value of the Darboux sum of every interval that contains x is less than or equal to the value of the Riemann integral of that interval.

#### sid9221

My lecture notes say:

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

$$L=\int_{a}^{b} f(x)=U$$

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

Never heard of a darboux integral will this definition suffice if asked for a riemann integral?

sid9221 said:
My lecture notes say:

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

$$L=\int_{a}^{b} f(x)=U$$

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.

sid9221 said:
My lecture notes say:

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

$$L=\int_{a}^{b} f(x)=U$$

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 $L(f, P)$, the sum of the minimum height and base of each interval, is 0 and each $U(f, P)$, 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.

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 $\displaystyle\sup_{P}L_{f,P} = \displaystyle\inf_{P}U_{f,P}$ iff $\forall \varepsilon>0 ~\forall P ~ \exists \delta>0~ \mbox{if} ~\lambda(P)<\delta ~\mbox{then} ~|U_{f,P}-L_{f,P}|<\varepsilon$, where $\lambda = \displaystyle\max_{1\le k \le n} |x_{k}-x_{k-1}|$

We supposed that $f$ is bounded.