How to prove that ##f(x,y)## is not integrable over a square?

In summary: I see. I think that ##2y<1## when ##y< \frac{1}{2}##, right? While ##2y>1## when ##y>\frac{1}{2}##. That it is because if ##y=\frac{1}{4}<\frac{1}{2}## then ##2y=\frac{1}{2}##, and if ##y=\frac{3}{2}>\frac{1}{2}## then...Yes, it is because ##2y<1## when ##y< \frac{1}{2}## and ##2y>1## when ##y>\frac{1}{2
  • #1
Davidllerenav
424
14
Homework Statement
Define ##f:R\subset\mathbb{R}^2\rightarrow\mathbb{R}## by
##f(x,y)=
\begin{cases}
2y & if \ x \ irrational \\
1 & if \ x\ rational
\end{cases}
##
Prove that ##f## is not integrable over the square ##R=[0,1]\times[0,1]##.
Hint: prove that for any partition ##P## of ##R##, ##L(f,P)<3/4## and ##U(f,P)>5/4##.
Relevant Equations
##L=\sum_{i,j}fm_{ij}A_{ij}##
##U=\sum_{i,j}fM_{ij}A_{ij}##
I'm confused with how Riemann sums work on double integrals. I know that ##L=\sum_{i,j}fm_{ij}A_{ij}## and ##U=\sum_{i,j}fM_{ij}A_{ij}## where ##m_{ij}## is the greatest lower bound and ##M_{ij}## is the least uper bound and ##A_{ij}## is the area of each partition.

##A_{ij}=\frac{1}{n^2}## for every partition. If I get it right, ##M_{ij}## must be where ##x_i## and ##y_i## are the maximum value on the square, so it must be at the the right top corner, while ##m_{ij}## must be at the bottom left corner. So ##M_{ij}=(\frac{i}{n},\frac{j}{n})## and ##m_{ij}=(\frac{i-1}{n},\frac{j-1}{n})##, right?

What should I do next?
 
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  • #2
Davidllerenav said:
Problem Statement: Define ##f:R\subset\mathbb{R}^2\rightarrow\mathbb{R}## by
##f(x,y)=
\begin{cases}
2y & if \ x \ irrational \\
1 & if \ x\ rational
\end{cases}
##
Prove that ##f## is not integrable over the square ##R=[0,1]\times[0,1]##.
Hint: prove that for any partition ##P## of ##R##, ##L(f,P)<3/4## and ##U(f,P)>5/4##.
Relevant Equations: ##L=\sum_{i,j}fm_{ij}A_{ij}##
##U=\sum_{i,j}fM_{ij}A_{ij}##

I'm confused with how Riemann sums work on double integrals. I know that ##L=\sum_{i,j}fm_{ij}A_{ij}## and ##U=\sum_{i,j}fM_{ij}A_{ij}## where ##m_{ij}## is the greatest lower bound and ##M_{ij}## is the least upper bound and ##A_{ij}## is the area of each partition.

##A_{ij}=\frac{1}{n^2}## for every partition. If I get it right, ##M_{ij}## must be where ##x_i## and ##y_i## are the maximum value on the square, so it must be at the the right top corner, while ##m_{ij}## must be at the bottom left corner. So ##M_{ij}=(\frac{i}{n},\frac{j}{n})## and ##m_{ij}=(\frac{i-1}{n},\frac{j-1}{n})##, right?

What should I do next?
It's not a matter of the extreme values of the coordinates for each of the squares of the partition. What matters is the extreme values taken on by the function, ##f##, for each of those squares.

On each of the squares of the partition, you can find some rational and some irrational values for ##x##. Then whether ##f## takes on a max or takes on a min, depends upon the comparison of ##2y## and ##1## .
 
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  • #3
SammyS said:
It's not a matter of the extreme values of the coordinates for each of the squares of the partition. What matters is the extreme values taken on by the function, ##f##, for each of those squares.

On each of the squares of the partition, you can find some rational and some irrational values for ##x##. Then whether ##f## takes on a max or takes on a min, depends upon the comparison of ##2y## and ##1## .
I don't understand how to I find the minimum and maximum of the function on each square. Can you explain it more please?
 
  • #4
Davidllerenav said:
I don't understand how to I find the minimum and maximum of the function on each square. Can you explain it more please?
It will depend on the value of y re whether 2y>1 or nor. Near x=1/2, there won't be much difference, but for smaller x , say x=1/10, the values will differ.
 
  • #5
Davidllerenav said:
I don't understand how to I find the minimum and maximum of the function on each square. Can you explain it more please?
On the square region, ##R##, the ##y## values range from 0 to 1, so ##2y## ranges from 0 to 2.
For any rational value of ##x, \ f## returns a value of 1, no matter the value of ##y##.
However, for any irrational value of ##x, \ f## returns a value which may be greater than 1, or may be lass than 1, depending upon the value of ##y##.

So, for what portion of ##R## is ##2y < 1\,?##

For what portion of ##R## is ##2y < 1\,?##
 
  • #6
WWGD said:
It will depend on the value of y re whether 2y>1 or nor. Near x=1/2, there won't be much difference, but for smaller x , say x=1/10, the values will differ.
@WWGD :
You have some typos there. (I think you mean for those x to be y instead. - or you should)
 
  • #7
SammyS said:
On the square region, ##R##, the ##y## values range from 0 to 1, so ##2y## ranges from 0 to 2.
For any rational value of ##x, \ f## returns a value of 1, no matter the value of ##y##.
However, for any irrational value of ##x, \ f## returns a value which may be greater than 1, or may be lass than 1, depending upon the value of ##y##.

So, for what portion of ##R## is ##2y < 1\,?##

For what portion of ##R## is ##2y < 1\,?##
I see. I think that ##2y<1## when ##y< \frac{1}{2}##, right? While ##2y>1## when ##y>\frac{1}{2}##. That it is because if ##y=\frac{1}{4}<\frac{1}{2}## then ##2y=\frac{1}{2}##, and if ##y=\frac{3}{2}>\frac{1}{2}## then ##2y=3##.
 
  • #8
Davidllerenav said:
I see. I think that ##2y<1## when ##y< \frac{1}{2}##, right? While ##2y>1## when ##y>\frac{1}{2}##. That it is because if ##y=\frac{1}{4}<\frac{1}{2}## then ##2y=\frac{1}{2}##, and if ##y=\frac{3}{2}>\frac{1}{2}## then ##2y=3##.
When you say, "and if ##y=\frac{3}{2}>\frac{1}{2}## then ##2y=3##", there at the end, I think you mean to say:

and if ##y=\frac{3}{4}>\frac{1}{2}## then ##2y=\frac32##.
 
  • #9
Davidllerenav said:
I see. I think that ##2y<1## when ##y< \frac{1}{2}##, right? While ##2y>1## when ##y>\frac{1}{2}##. That it is because if ##y=\frac{1}{4}<\frac{1}{2}## then ##2y=\frac{1}{2}##, and if ##y=\frac{3}{2}>\frac{1}{2}## then ##2y=3##.
And no matter how small your square, it will contain a rational x coordinate by density.
 
  • #10
A function is Riemann integrable iff it is bounded and continuous a.e.
 
  • #11
SammyS said:
When you say, "and if ##y=\frac{3}{2}>\frac{1}{2}## then ##2y=3##", there at the end, I think you mean to say:

and if ##y=\frac{3}{4}>\frac{1}{2}## then ##2y=\frac32##.
Why? Isn't ##\frac{3}{2}>\frac{1}{2}##?
 
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  • #12
WWGD said:
And no matter how small your square, it will contain a rational x coordinate by density.
Yes, but that complicates things, right?
 
  • #13
wrobel said:
A function is Riemann integrable iff it is bounded and continuous a.e.
So this function isn't bounded nor continuous, right? Because of the 2y.
 
  • #14
Davidllerenav said:
So this function isn't bounded nor continuous, right? Because of the 2y.

This post quotes a result that you are not able to use if you are trying to prove this from first principles, using Riemann sums. It was less than helpful, I would say.

That said, you have misinterpreted what the result is saying. If we have a theorem of the form:

"A function is integrable iff A and B hold", then if A and B both hold the function is integrable; and, if either of A or B does not hold, then the function is not integrable.

Moreover, in your case, the function is clearly bounded (by 2). The lack of continuity is the problem.
 
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  • #15
PeroK said:
This post quotes a result that you are not able to use if you are trying to prove this from first principles, using Riemann sums. It was less than helpful, I would say.

That said, you have misinterpreted what the result is saying. If we have a theorem of the form:

"A function is integrable iff A and B hold", then if A and B both hold the function is integrable; and, if either of A or B does not hold, then the function is not integrable.

Moreover, in your case, the function is clearly bounded (by 2). The lack of continuity is the problem.
Yes, I know that it doesn't help since I need to prove it. Oh, yes it is bounded, I forgot that y is bounded by 1, so the function is bounded by 2.
 
  • #16
loosely speaking, we have $$\int_Rfdxdy=\lim\sum_{i,j}f(\xi_i)\Delta x_i\Delta y_i,$$ where ##\xi_i=(x_i,y_i)## It remains to calculate this sum twice : for ##x_i\in\mathbb{Q}## and for ##x_i\notin\mathbb{Q}##
and make sure that the results will be different. It is it.
 
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  • #17
Davidllerenav said:
Why? Isn't \frac{3}{2}>\frac{1}{2}?

Okay, I've got an idea. Can you show first that for any partition:

##U(f, P) \ge 1##
 
  • #18
PeroK said:
Okay, I've got an idea. Can you show first that for any partition:

##U(f, P) \ge 1##
Using ##2y## as the upper bound?
 
  • #19
Davidllerenav said:
Using ##2y## as the upper bound?

What about using ##1## as the upper bound? There's a rational ##x## in every square.

It doesn't make showing that ##L(f, P) < 3/4## any easier, but at least it sorts out the upper bound.
 
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  • #20
Davidllerenav said:
Using ##2y## as the upper bound?

Okay, here's an alternative strategy.

1) You don't have to use every partition. As long as you choose a subset of partitions where the size of the intervals goes to 0 then that is enough to show non-integrability.

2) Choose the partitions ##4n##. This gives you clean boundaries in your sums, in particular at ##y = 1/4##.

3) Split ##L(f,p) = S_1 + S_2##, where ##S_1## is the sum on the bottom quarter (##y \le 1/4##) and ##S_2## is the area ##(y > 1/4)##

4) Show that ##S_2 \le 3/4##.

5) Show that ##S_1 \le 1/8##.

My suggestion is to understand this approach without going into the details first. Once you see what you are doing, try to translate it to the formal sums.

Note: another alternative is to show that ##f## is not integrable on the lower quarter only. That is enough to show that it is not integrable on a larger set.

In fact, that might be my preferred approach if I were doing the problem. Yes, definitely how I'd do it.
 
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  • #21
PeroK said:
Okay, here's an alternative strategy.

1) You don't have to use every partition. As long as you choose a subset of partitions where the size of the intervals goes to 0 then that is enough to show non-integrability.

2) Choose the partitions ##4n##. This gives you clean boundaries in your sums, in particular at ##y = 1/4##.

3) Split ##L(f,p) = S_1 + S_2##, where ##S_1## is the sum on the bottom quarter (##y \le 1/4##) and ##S_2## is the area ##(y > 1/4)##

4) Show that ##S_2 \le 3/4##.

5) Show that ##S_1 \le 1/8##.

My suggestion is to understand this approach without going into the details first. Once you see what you are doing, try to translate it to the formal sums.

Note: another alternative is to show that ##f## is not integrable on the lower quarter only. That is enough to show that it is not integrable on a larger set.

In fact, that might be my preferred approach if I were doing the problem. Yes, definitely how I'd do it.
Ok, I'll try to do it and I'll post my results and questions as soon as I wake up.
 
  • #22
It seems to me the problem is ill stated. It should specify that the function is not Riemann integrable. It certainly is Lebesgue integrable. In fact, it is the typical kind of example of a function that is Lebesgue integrable but not Riemann integrable.
 
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  • #23
And you require the set of discontinuities to have non-zero measure or uncountably-many discontinuities for Riemann integrability too. And, of course, boundedness.
 
  • #24
Davidllerenav said:
Why? Isn't ##\frac{3}{2}>\frac{1}{2}##?
Of course ##\frac{3}{2}>\frac{1}{2}##, but that has nothing to do with your problem.

The point ##(x,y)=\left(a, \dfrac{3}{2}\right), \text{where } 0\le a \le 1## is not in ##R##.
 
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  • #25
PeroK said:
Okay, here's an alternative strategy.

1) You don't have to use every partition. As long as you choose a subset of partitions where the size of the intervals goes to 0 then that is enough to show non-integrability.

2) Choose the partitions ##4n##. This gives you clean boundaries in your sums, in particular at ##y = 1/4##.

3) Split ##L(f,p) = S_1 + S_2##, where ##S_1## is the sum on the bottom quarter (##y \le 1/4##) and ##S_2## is the area ##(y > 1/4)##

4) Show that ##S_2 \le 3/4##.

5) Show that ##S_1 \le 1/8##.

My suggestion is to understand this approach without going into the details first. Once you see what you are doing, try to translate it to the formal sums.

Note: another alternative is to show that ##f## is not integrable on the lower quarter only. That is enough to show that it is not integrable on a larger set.

In fact, that might be my preferred approach if I were doing the problem. Yes, definitely how I'd do it.
One question, why do you choose ##4n## partitions? What do you mean by that?
 
  • #26
Davidllerenav said:
Hint: prove that for any partition ##P## of ##R##, ##L(f,P)<3/4## and ##U(f,P)>5/4##.
Relevant Equations: ##L=\sum_{i,j}fm_{ij}A_{ij}##
##U=\sum_{i,j}fM_{ij}A_{ij}##

I'm confused with how Riemann sums work on double integrals. I know that ##L=\sum_{i,j}fm_{ij}A_{ij}## and ##U=\sum_{i,j}fM_{ij}A_{ij}## where ##m_{ij}## is the greatest lower bound and ##M_{ij}## is the least upper bound and ##A_{ij}## is the area of each partition.
Those expressions for ##L## and ##U## are incorrect as I understand them.

Using your implied scheme of partitioning the square region, ##R##, into ##n^2## equal area sub-squares, let's define a general sub-square to be the region ##R_{ij} = [x_{i-1},x_{i}]\times [y_{j-1},y_{j}] ## .

(Note: This can also be expressed as: ##R_{ij} = \left[\frac{i-1}{n},\frac{i}{n}\right]\times \left[\frac{j-1}{n},\frac{j}{n}\right] ## . Also, it's helpful to have ##n## be even.)

Then let ##m_{ij}## be the greatest lower bound (GLB) of ##\{ f(x,y)\, : \, (x,y) \in R_{ij} \} ##, i.e. ##m_{ij} = \text{GLB}( f(R_{ij}) ) ## .

We then have ##\ \displaystyle L(f,P) = \sum_{i}^n \left(\sum_{j}^n m_{ij}\,A_{ij} \right)## .

Similarly, ##\ \displaystyle U(f,P) = \sum_{i}^n \left(\sum_{j}^n M_{ij}\,A_{ij} \right)##, where ##M_{ij} = \text{LUB}( f(R_{ij}) ) ##

(I think we have already established that if ##y \le \frac12 , \, \text{then } 2y \le 1 ## .)

If ##j\le\frac n 2 ## so that ##y_j \le \frac 1 2 ## , what are ##m_{ij}## and ## M_{ij}## ?
 
  • #27
Maybe this situation will be clearer.

Look at the case of n = 10, so that region ##R## is partitioned into 100 squares of equal size.

Consider the sub-square, ##R_{2,4}## . This is the set ##\{(x,\, y):\ 0.1 \le x \le 0.2 \text{, and } 0.3 \le y \le 0.4 \, \}##

Pick some rational ##x## in ##R_{2,4}##, such as ##x = 0.15 ##. What is the value of ##f(0.15,\, y)## for any ##y## such that ##0.3 \le y \le 0.4 \ ?##

Now, pick some irrational ##x## in ##R_{2,4}##, such as ##x =\sqrt{0.02\ }\approx 0.1414 ##. What is the range of values you get for ##f(\sqrt{0.02\ },\, y)## for any ##y## such that ##0.3 \le y \le 0.4 \ ?##

Now, what are the values for ##m_{2\,4} ## and for ##M_{2\,4} ## for this partition?
 

1. How do you define integrability for a function over a square?

Integrability over a square refers to the ability to calculate the definite integral of a function over a specific square region on a coordinate plane. This means that the function must have a finite area under the curve within the boundaries of the square.

2. What is the criteria for a function to be considered non-integrable over a square?

A function is considered non-integrable over a square if it has an infinite area under the curve within the boundaries of the square. This can occur if the function has vertical asymptotes, oscillates infinitely, or has a discontinuity within the square.

3. How can you prove that a function is not integrable over a square?

To prove that a function is not integrable over a square, you can use the Riemann sum method. This involves dividing the square into smaller sub-squares and calculating the sum of the areas under the curve within each sub-square. If the sum approaches infinity as the sub-squares get smaller, then the function is not integrable over the square.

4. Can a function be integrable over a square but not over a larger region?

Yes, a function can be integrable over a square but not over a larger region. This is because the boundaries of the larger region may include points where the function is non-integrable, even though it is integrable within the boundaries of the square.

5. Are there any alternative methods for proving non-integrability over a square?

Yes, there are other methods for proving non-integrability over a square, such as using the Cauchy criterion or the Lebesgue criterion. These methods involve analyzing the behavior of the function at specific points or within specific intervals to determine if it is non-integrable over the square.

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