0
$\begingroup$

This is the question below :

Problem.1) Prove that if $f : [a, b] \to \mathbb{R}$ is a bounded function, then $$ \underline{\int_{a}^{b}} f(x) \, \mathrm{d}x \leq \overline{\int_{a}^{b}} f(x) \, \mathrm{d}x. $$

This is my attempt at the question, I would like to get some guidance on how to go about doing this proof and the corresponding notation.

Attempted Solution.2)

$$ \text{For any partition } \Delta x = x_{i} - x_{i-1} $$

$$ \inf\left(f(x)\right) \leq \sup\left(f(x)\right) \ : \ x \in [x_{i-1}, x_{i}] $$

$$ m_{i} = \inf\{ f(x) : x \in [x_{i-1}, x_{i}]\} \leq M_i = \sup\{ f(x) : x \in [x_{i-1}, x_{i}]\} $$

$$ L(P, f) = \sum_{i=1}^{n} m_{i} \Delta x_{i} \leq U(P, f) = \sum_{i=1}^{n} M_{i} \Delta x_{i} $$

$$ \underline{\int_{a}^{b}} f(x) \, \mathrm{d}x = \sup\{ L(P, f) : P \in \mathcal{P}[a, b]\} \leq \overline{\int_{a}^{b}} f(x) \, \mathrm{d}x = \inf\{ U(P, f) : P \in \mathcal{P}[a, b]\}, $$

where $\mathcal{P}$ is a partition.

$$ \underline{\int_{a}^{b}} f(x) \, \mathrm{d}x \leq \overline{\int_{a}^{b}} f(x) \, \mathrm{d}x. $$


From the editor. Some typos are fixed by the editor, such as incorrect use of inf/sup and upper/lower sums, as well as sudden change in fonts.

$\endgroup$
  • $\begingroup$ Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. $\endgroup$ – Sangchul Lee Oct 4 '19 at 6:56
  • $\begingroup$ @SangchulLee thank you for the edit. $\endgroup$ – Amir Oct 4 '19 at 7:24
1
$\begingroup$

Your first steps show that $L(P,f) \leqslant U(P,f)$, where the same partition $P$ is used in the lower and upper sum. To finish you need to show that for (different) partitions $P$ and $Q$ we also have

$$\tag{*}L(P,f) \leqslant U(Q,f)$$

It then would follow that with $Q$ fixed,

$$\sup_P L(P,f) \leqslant U(Q,f),$$

and, subsequently,

$$\sup_P L(P,f) \leqslant \inf_QU(Q,f)$$

To prove (*) take a common refinement $R = P \cup Q$ and show that we must have

$$L(P,f) \leqslant L(R,f) \leqslant U(R,f) \leqslant U(Q,f)$$

I'll leave that to you with the hint that you should consider what happens to the ordering of lower and upper sums when a single new point is added to a partition.

$\endgroup$
  • $\begingroup$ I am confused as to why i need to consider different partitions since my question does not make mention of a second partition? $\endgroup$ – Amir Oct 4 '19 at 7:17
  • $\begingroup$ As I state in the answer you need this to show that any upper sum is an upper bound for the set of lower sums ranging over every partition. Then it follows the supremum of lower sums is less than or equal to this arbitrary upper sum. With that result you have the lower integral as a lower bound for the set of upper sums ranging over every partition, etc. $\endgroup$ – RRL Oct 4 '19 at 7:21
  • $\begingroup$ What you transcribed does not mention different partitions but that does not mean no consideration of different partitions is needed for a valid proof. $\endgroup$ – RRL Oct 4 '19 at 7:27
  • $\begingroup$ I know by taking a common refinement let us say on a upper sum would cause the upper sum to be reduced since an upper sum is an estimation and for a lower sum would cause the sum to increase. I am still confused as to the relationship between the different partitions and using the common refinement to show that the Lower sum <= to the upper sum in this instance. $\endgroup$ – Amir Oct 4 '19 at 7:30
  • 1
    $\begingroup$ You are on the right track now. Using the common refinement you get my last string of inequalities. Now you know that lower sum for ANY partition $P$ is less than or equal to upper sum for ANY partition $Q$. Please reread what I wrote. Starting with $L(P,f) \leqslant U(Q,f)$ first take the supremum of both sides with respect to $P$ and then take the infimum with respect to $Q$. $\endgroup$ – RRL Oct 4 '19 at 7:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.