But when we take the limit when n tends to infinity , the supremum we get is not supA
Not quite. We know the sup exists, so for every choice of partition the Darboux sum will converge to the sup.
I have a question about finding out if a function is integrable and what is the integral. My question is about taking a partition which divides every subinterval to equal length.
I will take an easy example. f(x)=x^{2 }, [0,b] , where b is a positive real number. I have uploaded a picture of what I did.
Now It is clear to me why it is integrable, because we have created squence of partitions , and U(f,Pn)-L(f,Pn)-> 0 when n tends to infinity.
Which means for every epsilon there exists N such that for every n>N we get U(f,P)-L(f,P)<epsilon which means that it's integrable.
Now my question is , we defined two sets , I will call them with other names so it won't be confusing.
A={U(f,P) : P is the partition of the interval [0,b]}
B={L(f,P) : P is the partition of the interval [0,b]}
supA=infB then f is integrable at the given interval.
But when we take the limit when n tends to infinity , the supremum and infimum we get is not supA and infB because we took the set of all partitions with its subintervals having same length. Now don't I have to show that the supremum that I have found is actually supA? or the infimum that I have found is infB? And if so, how can I do that? Or can I create two sets we will call them C and D
C={U(f,P) : P is the partition of the interval [0,b], every subinterval has the same length}
D={L(f,P) : P is the partition of the interval [0,b], every subinterval has the same length}
Since we have supC<supA<infB<infD (the notation here is acctually less or equal)
But since we have found supC=infD we actually come to the conclusion supC=supA=infB=infD
Can I do it like this? And if this is true, is it redundant? and if it is why is it redundant? finding out that U(f,P)-L(f,P)<epsilon is enough to come to the conclusion that the inf and sup or two sets are equal? Thank you.