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Given continuity prove a function is integrable.

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I am a student in advanced calculus and I am having an issue with a grade I just received. the question was as follows:

    If a function f:[0, 3]→ℝ is continuous use the Archimedes Riemann theorem to show that f is also integrable.

    I want to take my answer to my professor because I think I deserve more credit than I was given, which was half of the points. I actually think my answer deserves full credit, especially if the points were taken away based on the only comment that the professor wrote on the test. Before I bring this to him, I wanted some honest critiques from the forum because maybe I am missing something and he already dislikes me as it is, so I would rather not get any further on his bad side if I don't really have a case. I will provide you with my exact answer, along with his 5 word comment exactly as they were written by both of us. The question up top is already word for word, so to start off I did not think I was required to use the epsilon delta definition of continuity to prove integrability. Since he did not mention that I am assuming that was not a reason for any deduction.


    2. Relevant equations

    archimedes riemann theorem and the definition of uniform continuity.

    3. The attempt at a solution
    this is it, word for word:

    "If f is continuous over [0, 3], then by the extreme value theorem there is a partition P_n and a partition interval of [0, 3] which contains a u_k and v_k such that f(u_k)= m_k= inf of the interval, [x_(k-1), x_k], and f(v_k)=M_k= sup of [x_(k-1), x_k]. If we choose our u_k and v_k....."

    Here, he circled my u_k and v_k and wrote, "but you have already chosen u_k and v_k" and he drew a line to the first place I wrote it. This is the part I don't understand. It seems clear to me that I am referring to the same u_k and v_k that I named earlier. I called them by the same name, I used the word "our" to denote that I was referring to the same u_k and v_k from earlier in the proof, and I never said any words that would suggest they were different, like for instance "take some v_k and u_k" or "choosing another v_k and u_k". This is also the only place in the proof where he wrote anything. I will now continue exactly where I left off:

    "... and interval [x_(k-1), x_k] such that |M_k- m_k|= max |M_i - m_i|, then this statement is true: U(f,P_n)-L(f,P_n)= Ʃ (i=1 to n) [M_i- m_i][x_i- x_(i-1)]≤ |f(v_k)- f(u_k)|[Ʃ (i=1 to n)[x_i- x_(i-1)]= 3|f(v_k)- f(u_k)|. If we have chosen our partition interval P_n such that the limit as n→∞ of the gap(p_n)=0, since our v_k and u_k are contained in one of these intervals {v_n} and {u_n} are sequences such that limit as n→∞ of |v_n- u_n)| = 0. Since we know that a continuous function on a closed bounded interval is also uniformly continuous, limit as n→∞ of |v_n- u_n)| = 0 tells us that limit as n→∞ of |f(v_n)- f(u_n)| = 0
    as well. By the inequality given earlier, the fact that the limit as n→∞ of |f(v_n)- f(u_n)| = 0 ensures that the limit as n→∞ of [U(f,P_n)-L(f,P_n)]=0 as well. thus by the archimedes riemann theorem, f is integrable."

    Looking back, I guess I should have said that the limit proves there exists an archimedean sequence of partitions and hence blah blah blah, but i think this proof was pretty thorough, it is basically word for word from my text book. Out of 20 points, I received 10. Half credit. Now I want you guys to tear me apart and explain to me how this proof is flawed.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Nov 9, 2012
  2. jcsd
  3. Nov 9, 2012 #2
    i guess i also could have kept writing 3 beside the limits but again, this wasn't mentioned and even though I did disregard it, if the limit of something is zero as n --> inf, then 3 times that limit won't change anything.
     
  4. Nov 9, 2012 #3

    haruspex

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    I have to agree with your professor. You quote a theorem which guarantees the existence of some combination of entities, E, satisfying a given property P. The theorem only guarantees the existence of one such combination. You cannot then impose further conditions on the combination. The theorem does not say a combination satisfying the extra conditions exists.
    Not sure I'd have given even half credit, but I am a bit mean.
     
  5. Nov 9, 2012 #4
    Maybe I have nod idea what I am talking about, but I am pretty sure that to prove this one theorem, I only need to demonstrate that one such partition exists. my professor made that very clear. you can find a partition for any integrable function's domain that will make the condition limit as n→∞ of [U(f,P_n)-L(f,P_n)]=0 untrue. take f(x)= x² over [-1, 1]. If you took the partition {x_0= -1, 0, 1}, and then just took refinements that created new points in the interval [-1,0], like p_2={-1, -1/2, 0, 1}, p_3={-1, -1/2, -1/4, 0, 1}, p_4={-1, -1/2, -1/4, -1/8, 0, 1} and so forth, as n goes to ∞, the upper and lower sums will never match. if you can explain to me that i did not show at least one such partition exists, that is one thing, but I do not understand what extra conditions I have given that I haven't also proven.
     
  6. Nov 9, 2012 #5

    haruspex

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    Here's what you wrote in the OP:
    As I understand the EVT, it doesn't show existence of partitions. Rather, for any partition {[xk-1, xk]} you care to choose it says that for each interval [xk-1, xk] in that partition there will exist uk and vk such that f(uk)= mk= inf [xk-1, xk], etc.
    You cannot now 'choose' uk and vk such that anything. But maybe you didn't mean that. Perhaps you only meant to choose k, or choose the partition such that etc. Those variables you have control over. If so, the question becomes whether you can indeed choose such a k. If the partition is infinite then you may very well not be able to, so you need to say the partition is finite.
     
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