# Riemann-integrable functions

Hello,

can you provide me an example where the limits of a Riemann-integrable functiosn (or even continuous function may fail to be Riemann-integrable?

Thanks

CompuChip
Homework Helper
You mean: is there an example of a converging series $\{ f_n | n \in \mathbb Z \}$ of Riemann integrable functions of which the limit
$$f \binop{:=} \lim_{n \to \infty} f_n$$
is not Riemann integrable?

Here's a hint:

Do you know any of the "popular" functions that fail to be Riemann integrable? Why not try to construct a sequence of Riemann integrable functions that converge (pointwise) to this function.

You mean: is there an example of a converging series $\{ f_n | n \in \mathbb Z \}$ of Riemann integrable functions of which the limit
$$f \binop{:=} \lim_{n \to \infty} f_n$$
is not Riemann integrable?
I was referring to page 322 in Rudin.

So if you're given a fxn

f:[-2,3]->R defined by f(x) = 0 if x is rational and 4 is rational, then if is not RI, that is int(f(x),-2,3) DNE.

How does Lebesgue theory make it integrable?

HallsofIvy
Homework Helper
So if you're given a fxn

f:[-2,3]->R defined by f(x) = 0 if x is rational and 4 is rational, then if is not RI, that is int(f(x),-2,3) DNE.

How does Lebesgue theory make it integrable?
Did you mean "f(x)= 4 if x is not rational"?

Lebesque theory measures sets differently from Riemann theory, in a way that gives "measure" to a much larger collection of sets. In particular, in Lebesque measure any countable set (such as the rational numbers) has measure 0. Since the measure of the interval [-2, 3] has measure 3-(-2)= 5, just as in Riemann theory, and the disjoint union of the rational and irrational numbers give all of the interval, the measure of the set of irrational numbers in [-2, 3] is 5. If a function is constant on a measurable set, its integral over that set is that constant times the measure of the set. The integral of "f(x)= 0 if x is rational and f(x)= 4 if x is irrational" has integral 0(0)+ 4(5)= 20.

Of course, that has nothing to do with the original question.

CompuChip