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Riemann integrable

  1. Jun 14, 2005 #1
    hello all

    i just wanted to ask how would one prove that a function is riemann integrable through the definition that the lower integral has to equal the upper integral, an example on the function f(x)=x^2 would be of great help

  2. jcsd
  3. Jun 14, 2005 #2


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    Well, start off by writing exactly what you want to prove. (In formulas, I mean, not words)
  4. Jun 14, 2005 #3


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    this follows from the formula for the sum of the first n squares of integers.
  5. Jun 14, 2005 #4
    Here's a rough outline of what you would need to work through:

    Observe that [tex]f(x)=x^2[/tex] is strictly increasing on [tex] \left[ 0,\infty) [/tex], so for any subinterval [tex] \left[ x_{k-1},x_k \right) \epsilon \left[ 0,\infty \right) [/tex], you have
    [tex] m_k = f(x_{k-1}) = x_{k-1}^2[/tex]
    [tex] M_k = f(x_{k}) = x_{k}^2[/tex]
    [tex] \begin{align*}
    M_k (x_k-x_{k-1}) - m_k (x_k-x_{k-1}) \\
    &=(M_k - m_k)(x_k-x_{k-1}) \\
    &=(x_k^2-x_{k-1}^2)(x_k-x_{k-1}) \\
    &=(2 x_{k-1}(x_k-x_{k-1}) + (x_k-x_{k-1})^2)(x_k-x_{k-1}) \end{align}[/tex]
    where we are partioning an interval [a,b] with an arbitrary partition [tex]P=\{a=x_0<x_1<...<x_n=b\}[/tex]. To show that f(x) is Reimann integrable, you must show that the lower integral is equal to the upper integral, where
    [tex]L(f)=sup \{ \sum_{k=1}^n m_k(x_k-x_{k-1}) [/tex]​
    over all partitions P, and U(f) is defined parallel-y (as an infimum instead of a supremum). An easy way to get the inf/sup is to define your own sequence of partitions [tex](P_k)[/tex]: if it converges, then you can replace the inf/sup with the limit of the sequence of
    [tex] \sum_{k=1}^n m_k(x_k-x_{k-1}) [/tex]​
    defined over the [tex](P_k)[/tex]. For example try defining [tex]d=b-a[/tex] and [tex]P_i=\{a=x_0<x_1=a+\frac{d}{2^i}<x_1=a+2 \frac{d}{2^i}<...<x_{2^i}=b\}[/tex]. So in the k-th partition, [tex]x_i= i \frac{d}{2^i}[/tex]. Then figure out what the terms of [tex] | \left( L(f,P_k)-U(f,P_k) \right) | [/tex] look like, and determine if they converge; if they do, then you can show that [tex]U(f)-L(f)=0[/tex] (can you see how?)

    [tex] \begin{align*}
    \left( U(f,P_k)-L(f,P_k) \right) \\
    &=\sum_{j=1}^n (M_j-m_{j-1})(x_j-x_{j-1}) \\
    &=\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \end{align}

    You'll need to modify this proof slightly for the x<0 case (where is this assumption made?) and then figure out how to combine the results to show that [tex]f(x)=x^2[/tex] is Reimann-integrable over all of R.
    Last edited by a moderator: Jun 14, 2005
  6. Jun 14, 2005 #5


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    is it that hard? for x^2 on [0,a] you have a subdivision of length a/n, and the upper value on [(k-1)a/n, k/n] is (ka)^2/n^2, while the lower value is [(k-1)a]^2/n^2.

    so the lower sum is (a/n)[0^2/n^2 + a^2/n^2 +.......+[(n-1)a]^2/n^2],

    and the upper sum is (a/n)[a^2/n^2 + .....+[(n-1)a]^2/n^2 + [(n)a]^2/n^2].

    thus the difference between upper and lower sums is (a/n)[(n)a]^2/n^2 = a^3/n, which goes to zero as n goes to infinity. done.

    i'm a little sleepy but isn't this it?

    by the way this shows that what i said before is not needed. i.e. you only need that formula to evaluate the integral. existence is much easier.
    Last edited: Jun 14, 2005
  7. Jun 14, 2005 #6
    Well, yes. I just threw in a bit of extra rigor - our end results are identical.
  8. Jun 14, 2005 #7


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    extra rigor? where is my argument unrigorous?

    oh is ee,l you gave a lot of detail i assumed as obvious.
  9. Jun 14, 2005 #8
    Well, the OP asked for proof 'from defintion', so I thought to throw in , you know, how to go from a superemum over partitions to a limit of any convergent sequence of partitions, and then to choose a nice-looking sequence - it's stuff that wasn't obvious to me when I took first-year real analysis (lots of lemmas were involved, iirc).

    Your proof is plenty rigorous, no offence intended.
  10. Jun 14, 2005 #9


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    you are right. i am not as energetic as you are tonight. good job. peace
  11. Jun 15, 2005 #10
    so far i have worked my way through to get this
    [tex]U(f)-L(f) \le U(f,P_k) - L(f,P_k) = \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) [/tex]
    but i dont understand how how you could make up a sequence of partitions, to get this
    [tex]\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) <\epsilon[/tex]
    well especially the the sequence of partition you used i dont understand it, like what is a sequence of partitions isnt it just an ordered set of partitions?

    please help

  12. Jun 15, 2005 #11
    Consider the set of all partitions {P} of [a,b], defined as
    The [tex]x_k[/tex]s divide your interval into a countable number of sub-intervals [tex] \left[ x_{k-1}, x_k \right] [/tex]. The [tex]definition[/tex] of the lower/upper integral is rather abstract and useless - it defines them as the supremum/infimum of [tex]L(f,P)[/tex] and [tex]U(f,P)[/tex] respectively on the set of all possible partitions P. But there exists a nice lemma that if you take a sequence of partitions [tex](P_k)[/tex], and the limit of their partition sizes [tex]\delta = x_{k}-x_{k-1}[/tex] goes to zero (I think that's the requirement, it might be more general), then the supremum (infimum) of the [tex]L(f,P)[/tex] or [tex]U(f,P)[/tex] is the same as the limit of the sequence [tex]U(f,p_k)[/tex]:
    sup{[tex]L(f,P):P[/tex] is a partition of [a,b]}=lim [tex] U(f,p_k)[/tex]
    Intuitively, as your partition gets "finer and finer", the error of the lower (upper) sum gets less. I forgot how the rigorous version goes here. The point is, instead of working with supremums, you work with a limits of sequences (which are 'nicer').

    I'm not entirely sure what you're asking about - let's see, the [tex]\sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) [/tex]
    is unique to f(x)=x^2, under any partition (I derived at the top of my first post, assuming x>0 and that x^2 is strictly increasing; you get something very similiar when x<0).
    To actually work with this, it would be nice to actually choose a sequence of partitions (that gets succesively finer): an easy one is
    [tex]P_k=P \{ a=x_0<x_1=a+\frac{b-a}{k}<...<x_j=a+j\frac{b-a}{k}<...<x_k=a+k\frac{b-a}{k}=b \} [/tex]
    Which reduces down to what mathwonk was using: (I'll quote his post)
    lim \left(U(f)-L(f) \right)&=lim\left( U(f,P_k)-L(f,P_k) \right) \\
    &=lim \sum_{j=1}^n (2 x_{j-1}(x_j-x_{j-1}) + (x_j-x_{j-1})^2)(x_j-x_{j-1}) \\
    &=lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\
    &=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} \\
    &=0 \end{align}[/tex]
    Last edited by a moderator: Jun 15, 2005
  13. Jun 15, 2005 #12
    Hope this is making some sense...
  14. Jun 15, 2005 #13
    hello there

    I have to admit this sequence stuff certainly makes things look nice, or else I would have dwelled within the supremum and infimum stuff, well last of all i cant see how you made this step?

  15. Jun 15, 2005 #14
    From the definition of the particular partition sequence [tex](P_k)[/tex] I'm using,
    and so
    [tex]\begin{align*}&= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) + \left( \frac{b-a}{k} \right) ^2 \right) \left( \frac{b-a}{k} \right) \\
    &= lim \sum_{j=1}^{k}\left( 2x_j \left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\
    &= lim \sum_{j=1}^{k}\left( 2j\frac{b-a}{k} +2a\left( \frac{b-a}{k} \right) \right) \left( \frac{b-a}{k} \right) \\
    &=lim_{k \rightarrow \infty } ((k^2+k) +2a) \frac{(b-a)^3}{k^3} \\
    &=lim_{k \rightarrow \infty } (k^2 +2a) \frac{(b-a)^3}{k^3}\\
    &=lim_{k \rightarrow \infty } \frac{(b-a)^3}{k} + \frac{2a(b-a)^3}{k^3}\\
    &= 0 \end{align}[/tex]

    Apparently I messed up slightly - I copied mathwonk's result, and his interval was [0,b] and mine was [a,b], hence the extra term in the expressions - anyway it all goes to zero.

    edit: the "formula for the sum of the first n squares of integers" is how you get rid of the sum in the 5th line.
    Last edited by a moderator: Jun 15, 2005
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