## Thursday, August 7, 2008

### The Riemann condition

Continuing a topic I started a few posts ago, Definite Integrals, I want to discuss some properties of the definition of the integral according to Darbo, and to prove a theorem called Riemann condition.
This post builds heavily on the previous post on this subject, so it will be a good idea to read it before starting to read this post.

In this post f(x) is a bounded function on the interval [a,b] to the real numbers, and P is a division of the interval [a,b].

Firstly, lets show that the infimum of the Upper Darbo Sums (U(P,f)) is alway large or equal to the supremum of the Lower Darbo Sums (L(P,f)). To proof this it is enough to show that for any two divisions P1 and P2 the upper sum according to P2 is greater or equal to the lower sum according to P1. To do this we will need a simple lemma: "If all the points in P are also in P' than

$U(P,f)>U(P',f), L(P,f))<(P',f)$"

The inequalities are all weak. I will proof one of them and the second one can be proofed exactly the same. Lets suppose that there is only one extra point in P'. We can suppose that this new point is between the first two points in P. Than:

$L(P',f)-L(P,f)=\sum m'_{i}(x'_{i}-x'_{i-1})-\sum m_{i}(x_{i}-x_{i-1})=$
=$m'_{1}(x_{t}-x_{0})+m'_{2}(x_{1}-x_{t})-m_{1}(x_{1}-x_{0})$

Now, since m is the infimum of f on the segment x(i)-x(i-1) we get immediately that the result is greater than zero (or equal), as needed. Now by induction we can show that this is true for any number of extra points in P'. And with this the lemma is proven.

Now we can use this lemma. Lets look on P3 - the union of P1 and P2. According to the lemma:

$L(P1,f)

And this is exactly what I wanted to show.

Riemann condition
The next step is to prove the Riemann condition theorem. This theorem says that: "The function f(x) has an integral only and only if:
$\forall \epsilon>o \exist P$ $U(P,f)-L(P,f)<\epsilon$

In the first direction, because of the theorem I proofed above we get immediately that if the above condition is true than:

$infU(P,f)-supL(P,f)<\epsilon$

And since it is true for any epsilon, they must be equal. But if they are equal we get that f has an integral according to the definition. In the second direction, if we will suppose that the function has an integral than we will get that according tot he definition of infimum and supremum, for any epsilon large than zero, there is a P1 and P2 such that:

$L(P1,f)>supL(P,f)-\frac{\epsilon}{2}$ $U(P2,f)

Now all we need to do is to look on P3 - the union of P1 and P2. Than according to the theorem I proofed in the beginning of this post:

$L(P3,f)>supL(P,f)-\frac{\epsilon}{2}$ $U(P3,f)

Since the integral exists we can write than that:

$U(P3,f)-L(P3,f)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

And we are done prooving the Riemann condition.

In the next post on this subject I will show (and proof) another condition, which is very similar to Riemann condition but is more easy to work with.