Riemann Integrability of f(x) = x on [0,1]

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SUMMARY

The function f(x) = x for rational x and f(x) = 0 for irrational x on the interval [0,1] is not Riemann integrable. The reasoning involves taking an equipartition of n equal subintervals and calculating the integral as the limit of the sum, yielding a value of 1/2. However, since each subinterval contains an irrational number, the lower Riemann sum remains 0 for all partitions, leading to a contradiction. Thus, the conclusion is that the function is not Riemann integrable, although it is Lebesgue integrable with a value of 0.

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sihag
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f(x) = x , if x is rational
= 0 , if x is irrational
on the interval [0,1]

i just wanted to check if my reasoning is right.

take the equipartition of n equal subintervals with choices of t_r's as r/n for each subinterval.

calculating the integral as limit of this sum (and sending the norm to 0) i got 1/2 as my value.

now if f were to be R - integrable the value of the integral must be 1/2.
but each subinterval for any partition would contain an irrational so the lower R sum would be 0, for all partitions of [0,1]
this yields 0 as the value of the integral.

the two values contradict.
 
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Yes, if that function were integrable, then you would get two contradictory values. Conclusion: that function is not (Riemann) integrable.

(It is Lebesque integrable: since the rational numbers have measure 0, it Lebesque integral is 0. If the function were f(x)= x if is is irrational, 0 if x is rational, then its Lebesque integral would be 1/2.)
 

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