Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

A bounded function f:[a,b]->R is integrable on [a,b] if there exists L in R such that for any ϵ>0 , there exists Pϵ such that |S(P,f)−L|<ϵ for any partitions P⊃Pϵ Prove it!

OpenStudy (perl):

good luck :)

OpenStudy (anonymous):

help me to prove it @perl

OpenStudy (perl):

well, i can try

OpenStudy (perl):

what is L >

OpenStudy (anonymous):

L is real number

OpenStudy (anonymous):

\[L \in \mathbb{R} \]

OpenStudy (anonymous):

What exactly are you trying to prove? That is just the definition of Riemann integrability.

OpenStudy (perl):

im not sure

OpenStudy (perl):

this is a definition, you dont prove a definition :)

OpenStudy (perl):

i think he is trying to prove that a bounded function is integrable, oh

OpenStudy (anonymous):

this is not definition, it's a theorem..

OpenStudy (anonymous):

Definition : A bounded function f:[a,b] is integrable on [a,b] if there exists \[L \in \mathbb{R} \] such that for any \[\epsilon >0\] , there exists \[\delta >0\] such that for any partition P of [a,b] such that \[||P||>\delta \], we have \[|S(P,f)-L|<\epsilon \]

OpenStudy (perl):

Something is missing here. x-1/2 is integrable between 0 and 1, but it is not bounded.

OpenStudy (perl):

x^(-1/2)

OpenStudy (perl):

ok so you want to prove that a bounded function is Riemann integrable?

OpenStudy (perl):

there is also Darboux integrable

OpenStudy (anonymous):

no, i want to prove that theorem that using refinement of partition of P..

OpenStudy (perl):

I think its false. you can have an unbounded function that is integrable

OpenStudy (perl):

this doesnt look like a theorem, it looks like a definition of a bounded integrable function

OpenStudy (anonymous):

Okay, forget about that.. How to show that definition of Riemann integral and Darboux are equivalent?

OpenStudy (perl):

this is just the definition of riemann integrable , or riemann criterion

OpenStudy (anonymous):

Owh..thanks..:) I have next question : Show that definition of Riemann Integral and Darboux are equivalent.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!