Riemann-integrable functions

You mean: is there an example of a converging series [itex]\{ f_n | n \in \mathbb Z \} [/itex] of Riemann integrable functions of which the limit
[tex]f \binop{:=} \lim_{n \to \infty} f_n[/tex]
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.

 

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.

 

Personally I like that in measure theory, statements like
"If f is constantly equal to c almost everywhere on [a, b], then the integral of f over that interval is equal to (b - a) * c"
can be rigorously defined (once I'd be more clear about the measure). In particular, the statement "almost everywhere" has a well-defined meaning which usually corresponds to ones intuition (although admittedly, intuitively Q may seem larger than it is )