Checking for integrability on a half-open interval

In summary: The function is integrable because for every \epsilon>0 there exists a partition P such that U(f,P) - L(f,P) < \epsilon.
  • #1
schniefen
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Homework Statement
For a bounded, continuous and monotonous function on a half-open interval ##(a,b]##, how does one check if the function is integrable? (specifically Darboux integrable)
Relevant Equations
My definition of Darboux integrable: ##U(f,P)-L(f,P)<\epsilon## for all ##\epsilon>0##
For a closed interval ##[a,b]## I have learned that ##U(f,P)-L(f,P)=\frac{(f(b)-f(a))\cdot(b-a)}{N}## where ##N## is the number of subintervals of ##[a,b]## (if ##f## is monotonically decreasing, change the numerator of the fraction to ##f(a)-f(b)##). However, if the interval is half-open, then ##f(a)## is no longer defined. How does one go about this issue?
 
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  • #2
I assume you mean monotonic, not monotonous, though a dubious case can be made for synonimity.
 
  • #3
More seriously, I assume you have been presented with a definition of the integral for cases other than monotonic. Try working from that definition.
 
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  • #4
Yes, I mean monotone of course. The only definition of Darboux integrability I’ve been given is the epsilon one above, although this can be written somewhat differently. And when I check the integrability for a certain bounded, continues and monotone function on a closed interval I make use of the fact that the difference between the upper and lower Riemann sums create a column of height ##f(b)-f(a)## and width ##(b-a)/N##. Then finding an ##N## for all ##\epsilon>0## is quite easy.
 
  • #5
Just think about how to extend the argument slightly. You have monotonicity and continuity given.
 
  • #6
Could one write ##\lim\limits_{x \to a^+}f(x)## instead of ##f(a)##? Is this limit evaluable?
 
  • #7
schniefen said:
Problem Statement: For a bounded, continuous and monotonous function on a half-open interval ##(a,b]##, how does one check if the function is integrable? (specifically Darboux integrable)
Relevant Equations: My definition of Darboux integrable: ##U(f,P)-L(f,P)<\epsilon## for all ##\epsilon>0##

For a closed interval ##[a,b]## I have learned that ##U(f,P)-L(f,P)=\frac{(f(b)-f(a))\cdot(b-a)}{N}## where ##N## is the number of subintervals of ##[a,b]## (if ##f## is monotonically decreasing, change the numerator of the fraction to ##f(a)-f(b)##). However, if the interval is half-open, then ##f(a)## is no longer defined. How does one go about this issue?

Don't use that formula: it assumes a closed interval and is therefore not appropriate to a half-open interval.

In fact, [itex]f[/itex] will be Darboux integrable on [itex](a, b][/itex] because it is bounded and continuous; it is not necessary that it be monotonic (for example, [itex]\sin (x^{-1})[/itex] is integrable on [itex](0,1][/itex]).

The starting point is the definition of Darboux integrability:

A function [itex]f : (a,b] \to \mathbb{R}[/itex] is integrable if and only if for every [itex]\epsilon > 0[/itex] there exists a partition [itex]P[/itex] such that [itex]U(f,P) - L(f,P) < \epsilon[/itex].

The key observation is that as [itex]f[/itex] is bounded the contribution to [itex]U(f,P) - L(f,P)[/itex] from the first subinterval is bounded above by [itex]K\delta[/itex] where [itex]\delta[/itex] is the width of the subinterval and [itex]K \geq 0[/itex] does not depend on the choice of partition. This can be made arbitrarily small by suitable choice of [itex]\delta[/itex]. As [itex]f[/itex] is continuous and thus integrable on the closed bounded interval [itex][a + \delta, b][/itex], the contribution to [itex]U(f,P) - L(f,P)[/itex] from [itex][a + \delta, b][/itex] can be made arbitrarily small by suitable choice of subintervals.
 
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  • #8
pasmith said:
The key observation is that as [itex]f[/itex] is bounded the contribution to [itex]U(f,P) - L(f,P)[/itex] from the first subinterval is bounded above by [itex]K\delta[/itex] where [itex]\delta[/itex] is the width of the subinterval and [itex]K \geq 0[/itex] does not depend on the choice of partition. This can be made arbitrarily small by suitable choice of [itex]\delta[/itex]. As [itex]f[/itex] is continuous and thus integrable on the closed bounded interval [itex][a + \delta, b][/itex], the contribution to [itex]U(f,P) - L(f,P)[/itex] from [itex][a + \delta, b][/itex] can be made arbitrarily small by suitable choice of subintervals.
What is ##K##? If one were to prove that a piece-wise continuous function is integrable, say ##f(x)=\tan{(x)}## on ##[0,\pi/3)## and ##f(x)=\sin{(x)}## on ##(\pi/3,\pi]##, does one simply show that we can treat both intervals as closed intervals?
 
  • #9
What does bounded mean? That should give you a clue to what K might be.
 

What does it mean for a function to be integrable on a half-open interval?

Integrability on a half-open interval means that the function can be integrated (i.e. the area under the curve can be calculated) on the interval, excluding one of the endpoints.

How do you check for integrability on a half-open interval?

To check for integrability on a half-open interval, you can use the Riemann integration method or the Lebesgue integration method. Both methods involve breaking the interval into smaller subintervals and evaluating the function on each subinterval.

What are the necessary conditions for a function to be integrable on a half-open interval?

The necessary conditions for integrability on a half-open interval are that the function must be bounded and have a finite number of discontinuities on the interval. Additionally, the function must be Riemann or Lebesgue integrable on each subinterval.

Can a function be integrable on a half-open interval if it has an infinite number of discontinuities?

No, a function cannot be integrable on a half-open interval if it has an infinite number of discontinuities. The function must have a finite number of discontinuities for it to be considered integrable on the interval.

What is the significance of checking for integrability on a half-open interval?

Checking for integrability on a half-open interval is important because it allows us to determine if a function can be integrated on that specific interval. This is useful in many areas of mathematics and science, such as calculating areas and volumes, and in solving differential equations.

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