Positive integral implies bounded below?

In summary, the integral of f on [a,b] exists and is positive, but it takes on the value 0 infinitely many times in any subinterval J.
  • #1
kirstin.17
5
0

Homework Statement


The integral of f on [a,b] exists and is positive.
Prove there is a subinterval J of [a,b] and a constant c such that f(x) >= c > 0 for all x in J.

Hint: Consider the lower integral of f on [a,b]


Homework Equations





The Attempt at a Solution


I don't see how the hint helps. Obviously both the upper and lower integrals are greater than 0. I tried considering refinement partitions of J extending to [a,b] but that doesn't get me back to getting the actual function bounded.

Since the lower integral is greater than 0, the lower sum for any partition is greater than 0 as well... I just don't see how this is helping and how to go from the sum/integral back to the original function. Is there a different approach that could be taken? Any help? Thx :)
 
Physics news on Phys.org
  • #2
What about the Dirac delta distribution?

If f is indeed a function, then what can you say about the integral if no such subinterval J, positive constant c exists?
 
  • #3
Have no idea what Dirac Delta Distribution is... haven't learned that.
 
  • #4
Try proof by contradiction (Lower integral converges to THE integral as norm of partition goes to zero)
 
  • #5
Consider the characeristic function of the irrationals in [a,b] (f(x) = 1 if x is an irrational number in [a,b], and f(x) = 0 if x is a rational number in [a,b]). The (Lebesgue) integral of f on [a,b] exists and is b-a which is positive, but it takes on the value 0 infinitely many times in any subinterval J.

That was just for your information. You're probably dealing with the Riemann integral. Suppose there is no subinterval with that property. Then on every subinterval [c,d] for a < c < d < b, what's the minimum value f takes on? What is the lower Riemann sum for that partition? Then what is the lower integral, i.e. the supremum of these lower sums? This should contradict the sentence: "The integral of f on [a,b] exists and is positive."
 
  • #6
Well, the function is supposed to be positive, so the characteristic function in AKG's response does not really apply (ya it can be modified). I hope my response was plain enough and will get you started.

gammamcc
 
  • #7
Nowhere does it say that the function has to be positive.
 
  • #8
gammamcc said:
Well, the function is supposed to be positive ...

You can't assume that. (1) The intent of the problem is to prove that the function must be positive over some subinterval. (2) It isn't true. All you know is that the [itex]\int_a^b f(x)dx > 0[/itex]. For example, [itex]\int_{-\pi}^{\pi/2}\cos x dx[/itex] is positive (1) but the integrand is negative on [itex][-\pi,0)[/itex].

Note that the proposition is false as both AKG and I have given counterexamples. Some other constraints (e.g., continuity) must apply to make to proposition true.
 
Last edited:
  • #9
OK. It's just that one is supposed to conclude it's positive ..on an interval. OK? So your are over analyzing it as you throw Lebesgue theory at someone you know is asking a classical analysis question.
 

What does "positive integral" mean?

The term "positive integral" refers to any whole number that is greater than zero. It can also refer to any number that can be expressed as a product of positive whole numbers.

What does it mean for something to be "bounded below"?

When something is said to be "bounded below," it means that there is a lower limit or boundary that it cannot go below. This can also be thought of as the minimum value for a particular quantity.

Why does a positive integral imply being bounded below?

This is because a positive integral, by definition, cannot be less than zero. Therefore, any quantity that is expressed as a positive integral must also have a lower limit of zero, making it bounded below.

Can a positive integral be unbounded below?

No, a positive integral cannot be unbounded below. This is because, as mentioned earlier, a positive integral has a minimum value of zero, which serves as its lower limit or boundary.

What are some examples of quantities that follow the rule "positive integral implies bounded below"?

Some examples include the number of students in a classroom, the number of apples in a basket, and the number of pages in a book. All of these quantities are expressed as positive integers and have a lower limit of zero, making them bounded below.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
407
  • Calculus and Beyond Homework Help
Replies
3
Views
212
  • Calculus and Beyond Homework Help
Replies
20
Views
436
  • Calculus and Beyond Homework Help
Replies
2
Views
109
  • Calculus and Beyond Homework Help
Replies
26
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
583
  • Calculus and Beyond Homework Help
Replies
4
Views
243
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
Back
Top