Does continuity prove integrability?

[peripatein]

Hi,

Homework Statement

I am now asked to prove that f: [0,1]->[0,1] defined thus
f(0)=0 and f(x)=1/10ⁿ for every 1/2ⁿ⁺¹<x<1/2ⁿ for natural n,
is integrable.

Homework Equations

The Attempt at a Solution

Would it suffice to show that f is continuous? I.e. that lim x->0 f(x) = f(0) = 0, since as x->0 n->infinity?

 

[SammyS]

Hi,

Homework Statement

I am now asked to prove that f: [0,1]->[0,1] defined thus
f(0)=0 and f(x)=1/10ⁿ for every 1/2ⁿ⁺¹<x<1/2ⁿ for natural n,
is integrable.

Homework Equations

The Attempt at a Solution

Would it suffice to show that f is continuous? I.e. that lim x->0 f(x) = f(0) = 0, since as x->0 n->infinity?

You have all strict inequalities.

What is f(x) for x = 1/2ⁿ ?

How is this function continuous?

 

[peripatein]

For x=1/2^n f(x)=10^(ln2/lnx), if I am not mistaken.
For integrability f has to be continuous, doesn't it?
Alternatively I could demonstrate that f(x) is bound and monotone, couldn't I?

 

[SammyS]

For x=1/2^n f(x)=10^(ln2/lnx), if I am not mistaken.
For integrability f has to be continuous, doesn't it?
Alternatively I could demonstrate that f(x) is bound and monotone, couldn't I?

Then I take it that [itex]\displaystyle f(x)=\frac{1}{10^n}\ \text{ for }\ x\in\left(\frac{1}{2^{n+1}}\,,\ \frac{1}{2^n}\right]\ .[/itex]

This function is discontinuous at x = 2^-n, for all natural numbers, n .

However, all bounded piecewise continuous functions are Riemann integrable on a bounded interval.

What happens near x=0 ?

 

[peripatein]

F(x) would converge to zero by squeeze theorem, would it not?
But if 10^(lnx/ln2)<f(x)<10^(1+lnx/ln2) then f(x) is bounded and monotone in [0,1], so why could I not simply use that to show that f is integrable in this interval?

 

[SammyS]

F(x) would converge to zero by squeeze theorem, would it not?
But if 10^(lnx/ln2)<f(x)<10^(1+lnx/ln2) then f(x) is bounded and monotone in [0,1], so why could I not simply use that to show that f is integrable in this interval?

Well, yes, f(x) is integrable.

You can integrate it from 0 to 1 by setting up an infinite sum (infinite series) .

 

[peripatein]

I know that f is integrable. My question still stands, I believe - how shall I demonstrate that rigorously? Will it be by showing that it is bounded and monotone?

 

[SammyS]

I know that f is integrable. My question still stands, I believe - how shall I demonstrate that rigorously? Will it be by showing that it is bounded and monotone?

That seems like the thing to do.

 

[peripatein]

Is there a better way to go about demonstrating that? I am not sure which infinite series might serve me ideally. Could it be one whose general term is 1/2^n?

 

[SammyS]

Is there a better way to go about demonstrating that? I am not sure which infinite series might serve me ideally. Could it be one whose general term is 1/2^n?

Do you know how to set up a Riemann sum to represent an integral?

 

[peripatein]

Apparently not well. I'd appreciate some guidance.

 

[peripatein]

Actually, since the interval is 1, dividing it to n segments would yield 1/n. Wouldn't it? Am I in the right direction?

 

[HallsofIvy]

Actually, since the interval is 1, dividing it to n segments would yield 1/n.

"Yield 1/n" for what?

Wouldn't it? Am I in the right direction?

You said a couple of times "For integrability f has to be continuous, doesn't it?"

No, it doesn't. If you are going to prove "f is integrable" then you had better review what properties a function must have to be integrable. What theorems do you know about when a function is integrable?

 

[SammyS]

The title of your thread is: Does continuity prove integrability?

The answer to this question is, Yes!, but continuity is not necessary for integrability.

At any rate, the function you have here is not continuous, so the question asked in the title doesn't apply.​

Do you know how to set up a Riemann sum to represent an integral?

Apparently not well. I'd appreciate some guidance.

Actually, since the interval is 1, dividing it to n segments would yield 1/n. Wouldn't it? Am I in the right direction?

If you were to partition the interval [0, 1] into n equal length partitions, then, yes, the length of each partition would be 1/n.
However, it is not necessary to use equal length partitions, and for this function it would make much sense to do so.

Do you know what a graph of f(x) looks like?

f(x):

(n=0)

f(x) = 1, for x ∈ (1/2, 1]​

(n=1)

f(x) = 1/10, for x ∈ (1/4, 1/2]​

(n=2)

f(x) = 1/100, for x ∈ (1/8, 1/4]​

(n=3)

f(x) = 1/1000, for x ∈ (1/16, 1/8]​

(n=4)

f(x) = 1/10⁴, for x ∈ (1/32, 1/16]​

...​

 

[peripatein]

Alright, but even if I divided the area under 1/10^n into rectangles of height 1/10^n and length 1/2^n+1 - 1/2^n, I would get an infinite geometric series which would still not converge to the actual area! Should I then integrate that expression in order to obtain the actual area between 0 and 1?

 

[SammyS]

Alright, but even if I divided the area under 1/10^n into rectangles of height 1/10^n and length 1/2^n+1 - 1/2^n, I would get an infinite geometric series which would still not converge to the actual area! Should I then integrate that expression in order to obtain the actual area between 0 and 1?

What is that series? Why wouldn't it converge?

Actually it does converge to the area between 0 and 1 .

 

[peripatein]

But then the sum will be a1/1-q, where q is the ratio. Which will be equal to 10/19, which seems a bit too much for such a small area (as presented by Wolfram).

 

[peripatein]

Moreover, what about the right angle triangles between the curve and each of the rectangles?

 

[SammyS]

But then the sum will be a1/1-q, where q is the ratio. Which will be equal to 10/19, which seems a bit too much for such a small area (as presented by Wolfram).

That is the answer.

The area under the function in the interval 1/2 < x ≤ 1 is 1/2 .

10/19 is just 1/38 greater than 1/2 .

 

[peripatein]

You claim that 10/19 is the answer, and yet I still don't quite understand how that could possibly be! What about the triangles between each rectangle and the curve itself? Were they taken under consideration?

 

[SammyS]

You claim that 10/19 is the answer, and yet I still don't quite understand how that could possibly be! What about the triangles between each rectangle and the curve itself? Were they taken under consideration?

What triangles between each rectangle? It's all rectangles.

You came up with that answer, yet you think it's not the correct answer.

 

[peripatein]

Is it not a curve? The rectangles do not cover the entire area under the curve, as between each rectangle and the curve there remains a small area in the similitude of a triangle.

 

[SammyS]

Is it not a curve? The rectangles do not cover the entire area under the curve, as between each rectangle and the curve there remains a small area in the similitude of a triangle.

Look at the function as I described it earlier. The function is defined for all x on [0, 1] . The graph of the function is a set of horizontal line segments. When considering the area under the function, that area can made up (in an exact manner) as a set of rectangles with no gap between any of those rectangles.

Give an instance of any place on the interval [0, 1] where you need to use a triangle to find the area.

 

[SammyS]

   https://www.physicsforums.com/attachment.php?attachmentid=54809&stc=1&d=1358440853

The above is a log graph and a standard graph of your function.

(From WolframAlpha)