What's Example of Lebesgue Integrable function which is not Riemann Integrable?
There are plenty. Can you think of a characteristic function of a nonempty measurable set (of finite measure) that is discontinuous everywhere? Why will this do?
There's example 7.4 on page 145 -- in the limit, this is the classic example of a non-Riemann-integrable function.
But I don't understand why this will do.
I get annoyed when people refer to examples in specific books- do they expect everyone to have the book in front of them? But here, you don't even say what book!
The simplest example of a Lebesque integrable function that is not Riemann integrable is f(x)= 1 if x is irrational, 0 if x is rational.
It is trivially Lebesque integrable: the set of rational numbers is countable, so has measure 0. f = 1 almost everywhere so is Lebesque integrable and its integral, from 0 to 1, is 1.
But if no matter how we divide the interval from x= 0 to x= 1 into intervals, every interval contains both rational and irrational numbers: the "lower sum" is always 0 and the "upper sum" is always 1. As we increase the number of intervals to infinity, those do NOT converge.
Consider the following set $A= Q \cap [0,1]$. Where Q is the set of rational numbers of course. Now consider the characteristic function of $A$ denoted $X_A$ defined as follow: $X_A(x)=0$ when $x \in A$ and $x=0$ otherwise. Since this function is almost zero everywhere, then its Lebesgue integral is clearly 0. However, it is easily proved that $X_A$ is not Riemann integrable. As an argument but not proof to support this, the function is discontinuous at uncountable number of points.
Vignon S. Oussa
Could you give us another, more complicated example? It seems like the Dirichlet function is everywhere!
What kind of example do you want?? You can also have
Thanks! I'm looking for some where the set of discontinuities and the set of continuities are both of non-zero measure...Is that posible...I know of a type of Cantor set which has positive measure...are there others?
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