Can a Continuous Function Fail to be Riemann-Integrable?

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In summary, the conversation discussed the concept of Riemann integrable functions and how Lebesgue theory measures sets differently, allowing for a larger collection of integrable functions. The conversation also touched on the idea of functions being constant almost everywhere and how this can be rigorously defined in measure theory.
  • #1
Nusc
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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
 
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  • #2
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?
 
  • #3
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.
 
  • #4
CompuChip said:
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?

I was referring to page 322 in Rudin.
 
  • #5
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?
 
  • #6
Nusc said:
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.
 
  • #7
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 :smile:)
 

FAQ: Can a Continuous Function Fail to be Riemann-Integrable?

What is the definition of a Riemann-integrable function?

A Riemann-integrable function is a real-valued function that can be approximated by a sequence of simple functions, and its integral can be computed using the Riemann sum formula.

What are the necessary conditions for a function to be Riemann-integrable?

A function must be bounded and have a finite number of discontinuities in a closed interval in order to be Riemann-integrable.

How is the Riemann integral computed for a function?

The Riemann integral is computed by partitioning the interval into smaller subintervals, calculating the area of rectangles under the graph of the function within each subinterval, and taking the limit of these areas as the subintervals approach zero.

Can a function have an infinite number of discontinuities and still be Riemann-integrable?

No, a function cannot have an infinite number of discontinuities and still be Riemann-integrable. It must have a finite number of discontinuities in order for the Riemann integral to be well-defined.

How is the Riemann integral related to the concept of area under a curve?

The Riemann integral represents the area under the curve of a function over a given interval. It is the limit of Riemann sums, which approximate the area under the curve using rectangles. Therefore, it provides a way to calculate the exact area under a curve.

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