Regardless of the partition, the upper sums will be...
there is a continues and positive f in [0,1]
prove that g(x) is not integrabile in [0,1]
g(x)=f(x) for rational x
g(x)=-f(x) for irational x
why its not integrible?
because f the endless points of disconinuety?
my prof showed a function
g(x)=1 for rational x
g(x)=1/x for irational x
and he said that it is integrible
i dont know why the original is not integrible
the upper sum will be
the lower sum
i dont know if the supremum is actually is f(x) because its not an actual number.
same thing for the infinum
i just guessed because this is the only thing we've got.
but in order to prove that its not integrabile
their subtraction shoudnld be lowe then epsilon
dont know how to show that
i just remmember the words of my prof that said that in our couse dereclet function is not integrible.
but in other courses it is integrible and the integral is 1
so when TheChaz said that the result is 1
i got remmemebered the above remark.
regradring your question:
yes it is possible if f(x)=0,but i was given that its possitive so i cannot happen