After an hour of googling this as you suggested, I still can't find the discussions or proof you were referring to. Can you please either send the link you found or help me with this? I don't know how to even start it.
Prove that the function defined by:
f(x)={1 if x is rational, 0 if x is irrational}
is not integrable on [0,1]. Show that no matter how small the norm of the partition, ||P||, the Riemann sum can be made to have value either 0 or 1
Hmmm. I found a proof in about 15 seconds.
First, you need to understand what it means to be integrable: If a function is continuous on a given interval, it’s integrable on that interval.
Now, if you show that f(x) is discontinuous at any point in [0,1] then it is not integrable on the interval. This tidy proof shows that f(x) is discontinuous at every point in [0,1].
http://www-history.mcs.st-and.ac.uk/~john/analysis/Lectures/L14.html
I apologize for the confusion. Unfortunately, as an AI text-based bot, I don't have the ability to browse the internet or share direct links. However, I can certainly help you understand how to approach proving that the function f(x) is not integrable on [0,1] and that the Riemann sum can be made to have a value of either 0 or 1, regardless of the partition norm.
To prove that the function is not integrable on [0,1], we can use the fact that a function is integrable if and only if the upper and lower Riemann sums converge to the same value as the partition norm approaches zero. In other words, if the upper and lower Riemann sums have different limits for any partition norm, then the function is not integrable.
Let's suppose that f(x) is integrable on [0,1]. In that case, the upper and lower Riemann sums should converge to the same value as the partition norm approaches zero.
For this particular function, since the set of rational numbers and the set of irrational numbers are both dense in the interval [0,1], any subinterval of [0,1] will contain both rational and irrational numbers. Therefore, any partition of the interval [0,1] will have at least one subinterval that contains both rational and irrational numbers.
Now, consider the upper Riemann sum (U) and the lower Riemann sum (L) for this function on any partition. The upper Riemann sum is the sum of the function values at the upper endpoints of each subinterval, while the lower Riemann sum is the sum of the function values at the lower endpoints of each subinterval.
Since there exists at least one subinterval that contains both rational and irrational numbers, the upper Riemann sum for that subinterval will be 1, and the lower Riemann sum for that subinterval will be 0. This is because f(x) is defined to be 1 for rational numbers and 0 for irrational numbers.
Thus, there will always be a subinterval where the difference between the upper and lower Riemann sums is 1 (U - L = 1). This will be true for any partition, regardless of how small the norm of the partition (||P||) is chosen.
Therefore, we can conclude that the function f(x) is not integrable on [0,1], as the upper and lower Riemann sums do not approach the same value, and we can always find a subinterval where the difference is 1.
I hope this explanation clarifies the proof for you. If you have any further questions, please feel free to ask!