![Prove that Dirichlet's discontinuous function is not Reimann integrable. Please provide a
detailed proof.
Hint: The Dirichlet discontinuous function is defined as D(x)={(c when x is rational) and (d
when x is irrational)}.
Solution
hello. today in the calculus class that I AP (along with one other grad student) we
introduced Riemann integration and the other said something to the effect that for a function to
be integrable, it must not have "too many holes" (informally speaking), that is it must only be
discontinuous on a finite number of points. I countered from my memory of real analysis (which
may be faulty, I do stats, not a mathematician proper :S) a theorem that as long as the function's
discontinuities are countable set of points, even if they are infinite, the funcion is integrable. But
then I thought about the Dirichlet function, which is defined (one of many ways) as f(x)=x for
rational x, 0 for irrational x. Since the rationals are countable the function should be integrable if
the theorem I recalled is valid but the DIrichlet function is clearly not Riemann integrable (it is
Lebesgue integrable, but that is something else altogether). I think I am misstating the theorem- I
knew it had something to do with countability, can anyone help me out? THanks. 24.92.85.35
(talk) 01:21, 3 December 2011 (UTC) A function is Riemann integrable if and only if the set of
discontinuities has Lebesgue measure zero. Measure zero sets include finite sets, countable sets,
and even some uncountable sets (like the cantor set). The set of all irrational numbers in [0,1] has
lebesgue measure one. The dirichlet function is discontinuous on all the irrational numbers and
therefore NOT Riemann integrable. For the longest time everyone worked hard on this to figure
what exactly are the necessary and sufficient conditions on a function to be integrable. Riemann
came along and fixed up the integral (meaning defined it properly) and put it upon solid ground
and proved many of its properties from this new definition, which everyone had previously taken
for granted since Newton. He made the question well posed and hence the integral is named in
his honor. And then Henri Lebesgue was the first one who successfully answered this question
and also generalized the Riemann integral.71.211.145.44 (talk) 03:02, 3 December 2011 (UTC)
To be precise, a function on an interval [a,b] is Riemann integrable if and only if it is bounded
and the set of discontinuities has Lebesgue measure zero. 78.13.145.217 (talk) 22:54, 6
December 2011 (UTC)](https://image.slidesharecdn.com/provethatdirichletsdiscontinuousfunctionisnotreimannintegr-230326103706-dcf1aaff/85/Prove-that-Dirichlets-discontinuous-function-is-not-Reimann-integr-pdf-1-320.jpg)
Prove that Dirichlets discontinuous function is not Reimann integr.pdf
- 1. Prove that Dirichlet's discontinuous function is not Reimann integrable. Please provide a detailed proof. Hint: The Dirichlet discontinuous function is defined as D(x)={(c when x is rational) and (d when x is irrational)}. Solution hello. today in the calculus class that I AP (along with one other grad student) we introduced Riemann integration and the other said something to the effect that for a function to be integrable, it must not have "too many holes" (informally speaking), that is it must only be discontinuous on a finite number of points. I countered from my memory of real analysis (which may be faulty, I do stats, not a mathematician proper :S) a theorem that as long as the function's discontinuities are countable set of points, even if they are infinite, the funcion is integrable. But then I thought about the Dirichlet function, which is defined (one of many ways) as f(x)=x for rational x, 0 for irrational x. Since the rationals are countable the function should be integrable if the theorem I recalled is valid but the DIrichlet function is clearly not Riemann integrable (it is Lebesgue integrable, but that is something else altogether). I think I am misstating the theorem- I knew it had something to do with countability, can anyone help me out? THanks. 24.92.85.35 (talk) 01:21, 3 December 2011 (UTC) A function is Riemann integrable if and only if the set of discontinuities has Lebesgue measure zero. Measure zero sets include finite sets, countable sets, and even some uncountable sets (like the cantor set). The set of all irrational numbers in [0,1] has lebesgue measure one. The dirichlet function is discontinuous on all the irrational numbers and therefore NOT Riemann integrable. For the longest time everyone worked hard on this to figure what exactly are the necessary and sufficient conditions on a function to be integrable. Riemann came along and fixed up the integral (meaning defined it properly) and put it upon solid ground and proved many of its properties from this new definition, which everyone had previously taken for granted since Newton. He made the question well posed and hence the integral is named in his honor. And then Henri Lebesgue was the first one who successfully answered this question and also generalized the Riemann integral.71.211.145.44 (talk) 03:02, 3 December 2011 (UTC) To be precise, a function on an interval [a,b] is Riemann integrable if and only if it is bounded and the set of discontinuities has Lebesgue measure zero. 78.13.145.217 (talk) 22:54, 6 December 2011 (UTC)