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Consider the bounded function f : [0, 1] + R defined on the closed interval [0, 1] by 0 т f(x) = { 15 if x is irrational, if

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Answer #1

The function is integrable.

Firstly, notice that whatever the partition may be, the lower sum will always be zero since function attains 0 on irrational values, and we have at least one irrational point in any interval. So, the supremum of the lower sums is zero as well.

Now, we need to see if the infimum of the upper sums is zero as well, in which case it will equal the supremum of the lower sums, which is already zero as we have seen above. Now, let's take a partition {1/2,1/4,1/8,1/16,...,1/(2^n)}. Notice that as we increase n, the upper sum tends to zero. So, we have proved that the infimum of the upper sums is zero.

Now we know that if the infimum of the upper sums equals the supremum of the lower sums (when they exist, i.e. are finite) that is the value of the integral of the function. So, answer to the second part (b2) is zero.

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