Checking the integrability of a function using upper and lowers sums

[Adesh]

TL;DR

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 $\epsilon$ definition of integrals.

Please guide me.

 

[Delta2]

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.

 

[Adesh]

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.

 

[Delta2]

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?

 

[Adesh]

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.

 

[Delta2]

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.