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Proving Riemann Integration

  1. Nov 7, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Suppose that f(x) is a bounded function on [a,b]

    If M = sup(f) and m = inf(f), prove that -M = inf(-f) and -m = sup(-f). If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

    Let f be integrable on [a,b]. Prove that -f is integrable on [a,b] and if m>0, prove that 1/f is integrable on [a,b].


    2. Relevant equations

    So M is the least upper bound for f on [a,b] and m is the greatest lower bound for f on [a,b].

    3. The attempt at a solution

    So there's lots of stuff involved in this question. I'll start by trying to prove the first thing :

    If m = inf(f) and M = sup(f), prove that -m = sup(-f) and -M = inf(-f).

    So we know that : m ≤ f ≤ M

    Forgive me if I'm wrong, but this seems like a one liner? Simply multiplying by (-1) gives :

    -m ≥ -f ≥ -M so that -m is the least upper bound for -f and -M is the greatest lower bound for -f.

    Before I go any further I'd like to check if I'm not jumping the gun a bit there.
     
    Last edited: Nov 7, 2012
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  3. Nov 7, 2012 #2

    micromass

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    Yes, I think you are jumping the gun. You correctly deduced that

    [tex]-M\leq -f\leq -m[/tex]

    and this shows that -m is an upper bound of -f. But why is it a least upper bound?? How does that follow from the above inequality? Same for -M being the greatest lower bound.
     
  4. Nov 7, 2012 #3

    Zondrina

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    Whoops I mistyped a few things in my original post. I fixed them now. I meant for M to be the sup and m to be the inf.
     
  5. Nov 7, 2012 #4

    micromass

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    Please don't edit your original posts, it destroys the flow of the topic and is hard to read :wink:

    Anyway, my reply still holds, you have only proven that -m is an upper bound, but why is it the least upper bound? And why is -M the greatest lower bound?
     
  6. Nov 7, 2012 #5

    Zondrina

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    Now as for what you were saying.

    Since f is a bounded function on a nonempty set of real numbers, it has to have a least upper bound ( also a greatest lower bound ).

    So if L1 is any upper bound for -f, and L2 is any lower bound for -f, then :

    [itex]L_1 ≥ -m ≥ -f ≥ -M ≥ L_2[/itex]
     
  7. Nov 7, 2012 #6

    micromass

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    How did you obtain that inequality?? And why does that imply anything?

    You need to prove: if a is another lower bound of -f, then a≤-M. And same for m.
     
  8. Nov 7, 2012 #7

    Zondrina

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    Ah yes, so I want to show that -m is the least upper bound and -M is the greatest lower bound.

    Right now we already have that -m is an upper bound for -f and -M is a lower bound for -f.

    If L1 is any upper bound bound for -f, we want to show L1 ≥ -m. So :

    L1 ≥ -f
    L1 ≥ -m

    This seems too obvious? Showing it for the lower bound is going to be exactly the same so I'll focus on this.
     
  9. Nov 7, 2012 #8

    micromass

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    So you say: if [itex]L_1\geq -f[/itex], then [itex]L_1\geq -m[/itex]?? Can you clarify this? That doesn't seem obvious to me.
     
  10. Nov 7, 2012 #9

    Zondrina

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    Whoops, I had a brain hiccup there, forgot to do this :

    L1 ≥ -f
    -L1 ≤ f

    The rest of the algebra is obvious if what I'm thinking is correct. I believe I can sub m in for f here?
     
  11. Nov 7, 2012 #10

    micromass

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    OK, but can you write this out with some explanations instead of only writing down the inequalities??

    For example, what you could writes:
    We want to prove that -m is the greatest upper bound of -f. So let L be another upper bound of -f, this means: L≥-f. Multiplying by -1 yields f≤-L. Now .....

    Can you complete the above? Try to write using sentences instead of just symbols.
     
  12. Nov 7, 2012 #11

    Zondrina

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    Yessir, so.

    Now ... since we are given that m ≤ f ≤ M and that m is the infimum of f, we know f attains the value m on the interval [a,b]. That is, f = m at some point. So :

    -L1 ≤ m
    L1 ≥ -m

    Hence L1 is an upper bound for -f and -m is a least upper bound for -f.
     
  13. Nov 7, 2012 #12

    micromass

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    Why should f attain the value m somewhere on the interval [a,b]??
    If you're using the extreme value theorem, then think again: it only holds for continuous functions.
     
  14. Nov 7, 2012 #13

    Zondrina

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    I just re-read the theorem, you're right. I haven't used it in so long I had forgotten. Continuity would've made things nice :(.

    So I need a different explanation, but the rest of what I said looks good.

    Now I argued that L1 was an upper bound for -f, so it would be a lower bound for f. Since m is the greatest lower bound for f, we get :

    -L1 ≤ m
    L1 ≥ -m

    Hence L1 is an upper bound for -f and -m is a least upper bound for -f.
     
  15. Nov 7, 2012 #14

    micromass

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    OK, that is good! (although the "Hence L1 is an upper bound of -f" is unnecessary since that was the hypothesis)
     
  16. Nov 7, 2012 #15

    Zondrina

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    Sweet, okay. I'll write everything nice and clean with a good explanation here then.

    So we know that m ≤ f ≤ M which implies that -m ≥ -f ≥ -M. What we want to show that -m is the supremum of -f and that -M is the infimum of -f. Right now we have that -m is an upper bound for -f and -M is an lower bound for -f.

    So suppose that L1 is any upper bound for -f. So we get L1 ≥ -f which yields -L1 ≤ f. Now, since we argued that L1 is any upper bound for -f, so it must be a lower bound for f. Since m is the greatest lower bound for f, we get -L1 ≤ m which implies that L1 ≥ -m. Hence -m is the least upper bound for -f.

    Suppose now that L2 is any lower bound for -f. So we get L2 ≤ -f which yields -L2 ≥ f. Now, since we argued that L2 is any lower bound for -f, it must be an upper bound for f. Since M is the least upper bound for f, we get -L2 ≥ M which implies that L2 ≤ -M. Hence -M is the greatest lower bound for -f.
     
  17. Nov 7, 2012 #16

    Zondrina

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    Now I'm going to attempt the second part of the question.

    If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

    So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M.

    This looks like an exact copy and paste of my post above except with a few different numbers if I'm not mistaken?
     
  18. Nov 7, 2012 #17

    micromass

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    OK, so that proves that 1/m is an upper bound of 1/f and that 1/M is a lower bound. So that implies indeed that 1/f is bounded. (where did you use that m>0 anyway?)

    But you still have more work to do if you wand to show that 1/M is the greatest lower bound and that 1/m is the least upper bound. The work is very analogous to above though.
     
  19. Nov 7, 2012 #18

    Zondrina

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    I'm assuming that M can't be zero either otherwise we would have a problem here, but here goes my attempt :

    So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M hence 1/f is bounded above by 1/m and bounded below by 1/M. We want to show that 1/m is the supremum of 1/f and 1/M is the infimum of 1/f. Right now we have that 1/m is an upper bound for 1/f and 1/M is a lower bound for 1/f.

    So suppose that Q1 is any upper bound for 1/f. Then we have Q1 ≥ 1/f which implies that 1/Q1 ≤ f. Now since Q1 is any upper bound for 1/f, 1/Q1 must be a lower bound for f. Now, since m is the greatest lower bound for f, we have that 1/Q1 ≤ m which yields Q1 ≥ 1/m. Hence 1/m is the least upper bound for 1/f.

    Now suppose that Q2 is any lower bound for 1/f. Then we have Q2 ≤ 1/f which implies that 1/Q2 ≥ f. Now since Q2 is any lower bound for 1/f, 1/Q2 must be an upper bound for f. Now, since M is the least upper bound for f, we have that 1/Q2 ≥ M which yields Q2 ≤ 1/M. Hence 1/M is the greatest lower bound for 1/f.

    I believe that should do it.
     
  20. Nov 7, 2012 #19

    micromass

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    That's correct! And you have written that perfectly!
     
  21. Nov 7, 2012 #20

    Zondrina

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    Whoop :D, so I only have one part left here : Let f be integrable on [a,b]. Prove that -f is integrable on [a,b] and if m>0, prove that 1/f is integrable on [a,b].

    So I'll be using some notation here, but I'll try to make it clear what is what.

    I have a definition here : If I = sup(sp) = inf(Sp) = J, then f is integrable.

    Now, [itex]s_p = \sum_{i=1}^{n} m_i Δx_i[/itex] and [itex]S_p = \sum_{i=1}^{n} M_i Δx_i[/itex].

    To fill in the last few blanks, [itex]m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex] and [itex]M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}[/itex]

    I'll start writing my proof in the next post since this is pretty cluttered as is.
     
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