Integrability
f ( x ) = ⎩ ⎪ ⎨ ⎪ ⎧ 0 if x is rational 1 if x is irrational
Is the function above Riemann integrable on
[
0
,
1
]
?
Is it
Lebesgue integrable
on
[
0
,
1
]
?
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1 solution
Great work! I'd also note that the Lebesgue integral is 1. Intuitively, this is because almost all of the numbers are irrational, so we have f ( x ) = 1 across almost the whole domain.
More formally, because the rational numbers are countable , and the measure of a countable union of disjoint sets is the sum of the measure of each, μ ( x ∈ [ 0 , 1 ] ∩ Q ) = 0 . And since μ ( [ 0 , 1 ] ) = 1 , we must have μ ( x ∈ [ 0 , 1 ] \ Q ) = 1 − 0 = 1 . Thus, the Lebesgue integral is 0 × 0 + 1 × 1 = 1 .
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μ(x∈[0,1]∩Q)=0, can you explain it?
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See https://math.stackexchange.com/questions/889633/measure-of-set-of-rational-numbers
Same as @afsheen ayub , can you explain μ(x∈[0,1]∩Q)=0? What is the union you are referring to, it seems like an intersection.
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Ah, I see you are referring to the union of the set where f(x)=1 and the set where f(x)=0.
See https://math.stackexchange.com/questions/889633/measure-of-set-of-rational-numbers
I answered No, Yes e.g. Not Riemann integrable and Lebesgue integrable, which the website said was wrong????
Anther way of viewing could be in the definition itself.As mentioned by Samir Khan,the function is a simple function. If we consider the Rienmann definition,we observe that there are infinitely many domain elements for the given function.This means that the sum cannot be closed under the Rienmann definiton. However as per Lebesgue's defintion,if we divide along the range,we see there are only two possible values of the range for the rational (R) and irrational (C) sets respectively.Thus we obtain a bounded form the function,making it integrable
Relevant wiki: Lebesgue Integration
The function is not Riemann integrable because the upper and lower sums do not converge, since every interval contains both infinitely many rationals and irrationals. The function is Lebesgue integrable however, because it is a simple function.