- #1

Adesh

- 735

- 191

- 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:

$$ sup \{L (f,P) : \text{P belongs to the set of partitions} \} = inf\{ U(f,P): \text{P belongs to the set of of partitions}\}$$

(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.

Please guide me.

*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.

Please guide me.