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

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 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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SammyS

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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## .
 
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?
 

WWGD

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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.
 

SammyS

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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\,?##
 

SammyS

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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)
 
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##.
 

SammyS

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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##.
 

WWGD

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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.
 
A function is Riemann integrable iff it is bounded and continuous a.e.
 
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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And no matter how small your square, it will contain a rational x coordinate by density.
Yes, but that complicates things, right?
 
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.
 

PeroK

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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.
 
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.
 
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.
 

PeroK

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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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PeroK

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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.
 
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.
 

Orodruin

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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.
 

WWGD

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And you require the set of discontinuities to have non-zero measure or uncountably-many discontinuities for Riemann integrability too. And, of course, boundedness.
 

SammyS

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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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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?
 

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