Checking the integrability of a function using upper and lowers sums

• I
In summary, the conversation discusses the topic of integrability in mathematics, specifically the conditions for a function to be integrable on a given interval. The conversation also includes a discussion on the concept of infimum and how it relates to the upper sum of a function on a partition. The conclusion is that the infimum is equal to zero, which is proven by considering the properties of infimum and using a proof by contradiction.
TL;DR Summary
A function is integrable if and only if ##sup\{L(f,P)\}=inf\{U(f,P)\}##.
Hello and Good Afternoon! Today I need the help of respectable member of this forum on the topic of integrability. According to Mr. Michael Spivak: A function ##f## which is bounded on ##[a,b]## is integrable on ##[a,b]## if and only if
$$sup \{L (f,P) : \text{P belongs to the set of partitions} \} = inf\{ U(f,P): \text{P belongs to the set of of partitions}\}$$
(notaions might cause some problem, ##L(f,P)## means the "lower sum of ##f## on the partition ##P##" and similarly for the ##U(f,P)##)
Now, let's consider a function ##f## which is bounded and defined on ##[0,2]## as
$$f(x)= \begin{cases} 0 & x\neq 1 \\ 1 & x =1 \end{cases}$$
Let's take a partition ##P## of ##[a,b]## as ## P = \{t_0, t_1, ... t_{j-1}, t_{j}, ... t_n\}##, such that ##t_{j-1} \lt 1 \lt t_j##. Define lower and upper sum on this partition,
$$L(f,P) = \sum_{i=1}^{j-1} m_i (t_i - t_{i-1}) + m_j ( t_j - t_{j-1}) + \sum_{i=j+1}^{n} m_i (t_i - t_{i-1})$$
##m_i## represents the minimum value of ##f## in the ##i## th interval, so essentially we have ##L(f,P) = 0## for all partitions.

$$U(f,P) =\sum_{i=1}^{j-1} M_i (t_i - t_{i-1}) + M_j ( t_j - t_{j-1}) + \sum_{i=j+1}^{n} M_i (t_i - t_{i-1})$$
##M_i## represents the maximum value of ##f## in the ##i## th interval, only ##M_j=1## while all the other ##M_i## are zero. So, we have ##U(f,P) = t_j - t_{j-1}## for the current partition that we have.

But we can make our partitions only in two ways with regards to ##1##, either ##1## will fall in between in some ##j## th interval or it will be an end point of two intervals. For the first case we found our upper sum as stated above, let's do the second case. Define the partition ##P' = \{t_0, t_1, ... t_{j-1}, 1 , t_{j+1}, ... t_n\}##, now the upper sum on this partition is
$$U(f,P') = \sum_{i=1}^{j-1} M_i (t_i - t_{i-1}) + M_j (1 - t_{j-1}) + M'_{j-1} (t_{j+1} - 1) + \sum_{i= J+2}^{n} M_i (t_i - t_{i-1})$$
We have ##M_j = M'_j = 1##, and all other ##M_i## are zero, therefore, for this case we have ##U(f,P) = t_{j+1} - t_j##.

Now, we can see that ##sup\{L(f,P)\} =0## but ##inf\{U(f,P)\} = ?##. I have no reason to conclude that the infimum of upper sum is zero and without it I cannot say ##f## is integrable although we know that ##f## is integrable by ##\epsilon## definiton of integrals.

You have already showed that $$U(f,P)=t_{j+1}-t_j$$ or $$U(f,P)=t_j-t_{j-1}$$ so it will be (in the first case but similar for the second case)$$inf(U(f,P))=inf(t_{j+1}-t_{j})$$. This is essentially equivalent to $$inf\{(x-y)|x>y>0\}=0$$ , because for any given partition P we can construct a partition P' that is the same as P , except that we ll choose ##t'_j## and ##t'_{j+1}## to be closer to each other than ##t_j## and ##t_{j+1}## are.

Delta2 said:
This is essentially equivalent to

$$inf{(x−y)|x>y>0}=0$$​
Sir, how you got the above infinimum equal to zero? Is it some property of infinimum, if yes then where can I learn about them? It’s just in Real Analysis that I find the words “sup” and “inf” and hence all I know is the manual process of finding the infinimum and supremum.

Sir, how you got the above infinimum equal to zero? Is it some property of infinimum, if yes then where can I learn about them? It’s just in Real Analysis that I find the words “sup” and “inf” and hence all I know is the manual process of finding the infinimum and supremum.
You can prove that the above infimum is zero by first noticing that ##x-y>0## so 0 is a lower bound, so ##infimum\geq 0## and then by arguing that if the infimum is some ##\epsilon>0## you could find ##x>0## and ##y>0## such ##(x-y)<\epsilon##.

But what exactly do you mean by the "manual process" of finding the infimum and supremum?

Delta2 said:
You can prove that the above infimum is zero by first noticing that ##x-y>0## so 0 is a lower bound, so ##infimum\geq 0## and then by arguing that if the infimum is some ##\epsilon>0## you could find ##x>0## and ##y>0## such ##(x-y)<\epsilon##.

But what exactly do you mean by the "manual process" of finding the infimum and supremum?
Manual process means taking out the minimum value from the set by inspection. But I think ##inf## need not to be in the set itself.

Manual process means taking out the minimum value from the set by inspection. But I think ##inf## need not to be in the set itself.
yes that's right, when the minimum exists (inside the set) then it is also the infimum, but sometimes the minimum does not exist, but the infimum exists and this is the case here.

1. What is the purpose of checking the integrability of a function using upper and lower sums?

The purpose of checking the integrability of a function using upper and lower sums is to determine if a function is integrable, or if it can be represented by a definite integral. This is important in calculus and real analysis, as it allows us to calculate the area under a curve and solve other important problems.

2. How do upper and lower sums differ from Riemann sums?

Upper and lower sums are similar to Riemann sums, but they use different methods to approximate the area under a curve. Riemann sums use rectangles of equal width, while upper and lower sums use rectangles of varying widths to better approximate the shape of the curve.

3. Can a function be integrable if it does not have a continuous derivative?

Yes, a function can still be integrable even if it does not have a continuous derivative. This is because the concept of integrability is based on the behavior of the function as a whole, not just its derivative. However, having a continuous derivative does make it easier to determine integrability using upper and lower sums.

4. How do you use upper and lower sums to determine if a function is integrable?

To determine if a function is integrable using upper and lower sums, you first divide the interval of integration into smaller subintervals. Then, you calculate the upper and lower sums for each subinterval and see if they converge to the same value. If they do, then the function is integrable. If they do not, then the function is not integrable.

5. Are there any limitations to using upper and lower sums to check integrability?

Yes, there are limitations to using upper and lower sums to check integrability. These methods only work for functions that are bounded and defined on a closed interval. Additionally, they may not be accurate for highly oscillating functions or functions with discontinuities.

• Calculus
Replies
2
Views
1K
• Calculus
Replies
7
Views
1K
• Calculus
Replies
16
Views
3K
• Calculus
Replies
3
Views
1K
• Calculus
Replies
1
Views
529
• Calculus
Replies
23
Views
3K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
1
Views
2K
• Calculus
Replies
1
Views
283
• Differential Equations
Replies
1
Views
990