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Lower integrals

  1. Jun 15, 2005 #1
    hello all

    another simple question would it be true to say

    [tex]\int_{a_{L}}^{b} f = lim_{k \rightarrow \infty }L(f,P_k)[/tex]

  2. jcsd
  3. Jun 16, 2005 #2
    I'm not sure what you mean, is this a continuation of the other thread? I strongly suggest you work through the relevant section of a Real Analysis text in rigor, paying attention to how things are defined and how things are proved..

    A Lower Sum of f(x) on [a,b], with the partition [tex]P= \{ a=x_0<x_1<x_2<...<x_n=b \}[/tex]:
    [tex]L(f,P)=\sum_{k=1}^{n} m_k \left( x_k-x_{k-1} \right) [/tex]
    where [tex]m_k[/tex] is a lower bound of f(x) on [tex] \left[ x_{k-1}, x_k \right] [/tex] (i.e., [tex]m_k \leq f(x) \forall x\epsilon \left[ x_{k-1}, x_k \right] [/tex].
    An Upper Sum of f(x) on [a,b] is defined
    [tex]U(f,P)=\sum_{k=1}^{n} M_k \left( x_k-x_{k-1} \right) [/tex]
    where [tex]M_k[/tex] is an upper bound of f(x) on [tex] \left[ x_{k-1}, x_k \right] [/tex] (i.e., [tex]m_k \geq f(x) \forall x\epsilon \left[ x_{k-1}, x_k \right] [/tex].

    The Lower Integral of f(x) on [a,b]:
    [tex]L(f) = sup \. L(f,P)[/tex] over all partitions of [a,b].
    The Upper Integral of f(x) on [a,b]:
    [tex]U(f) = inf \. U(f,P)[/tex] over all partitions of [a,b].

    The Reimann Integral: If U(f)=L(f), then [tex]\int_{a}^{b}f(x)dx[/tex] exists and is defined as

    Back to your question:

    You haven't defined everything. What you have is certainly not true - your [tex] \left P_k \right[/tex] looks like an arbitrary sequence, and you have no restrictions on f.
    Here's what is sufficient (you may have been thinking of): If f is bounded, and you have a sequence [tex] \left P_k \right[/tex] such that
    [tex] lim_{k \rightarrow \infty}\left( L(f,P_k)-U(f,P_k) \right)=0[/tex], then you can see from the definition that [tex]L(f)=U(f)=lim_{k \rightarrow \infty}L(f,P_k)=lim_{k \rightarrow \infty}U(f,P_k)[/tex].
    The kind of sequence of partitions that might satisfy this involves partitions getting 'finer and finer'; for example, where each partition is a refinement of the previous one.
    If you look at a very clean, sterilized example, like f(x)=x^2 which is continuous, bounded, and has a bounded derivative, than any partition sequence where the partition size approaches zero will be sufficient. To prove this, given a max. partition size [tex]\delta[/tex], find an upper bound on [tex]M_k-m_k[/tex] on an arbitrary subinterval (hint: bounded derivative) - and plug this into the sums to show that the upper and lower sums converge to the same thing.
    This is NOT a generally applicable procedure; see what happens if f is unbounded or very discontinuous.

    (edited for errors)
    Last edited by a moderator: Jun 16, 2005
  4. Jun 16, 2005 #3
    hello rachmaninoff

    thanxs for the your explanation it really helped to link the concepts together on what i am studying, im thinking that if k determines the number of sub-intervals(thats what k is, am i right?) then as k goes to infinity then the partition gets finer and finer and so the limit of L(f,Pk) and U(f,Pk) approaches the lower integral and the upper integral respectively as long as f is bounded and that pk is defined by a particular sequence, would i be correct, these are the thoughts that came to me from our other thread where you said

    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.

    when you explained finer and finer do you mean that the difference between any 2 consecutive elements will get smaller and smaller as the number of elements in the partition increases ?

  5. Jun 16, 2005 #4
    Here's a pathological case that might explain your questions:

    Consider the sequence of partitions of the interval [0,1] where
    [tex]P_1= \left( \left[ 0,\frac{1}{2} \right] , \left[ \frac{1}{2},1 \right] \right) [/tex]
    [tex]P_2= \left( \left[ 0,\frac{1}{2} \right] , \left[ \frac{1}{2},\frac{3}{4} \right] , \left[ \frac{3}{4},1 \right] \right) [/tex]
    [tex]P_3= \left( \left[ 0,\frac{1}{2} \right] , \left[ \frac{1}{2},\frac{3}{4} \right] , \left[ \frac{3}{4},\frac{7}{8} \right] , \left[ \frac{7}{8},1 \right] \right) [/tex]

    and in general

    [tex]P_n=\left{ x_0=0<x_1=\frac{1}{2}<...<P_k=\frac{2^{i-1}}{2^i}<....<x_n=1 \right}[/tex]

    If you take your f(x)=x^2 example and apply the lower/upper sums to it using these partitions, you will find that they do not converge to the correct values (the lower/upper integrals) at all!

    The definition of Lower/Upper Integral in terms of the supremum(infimum) of the lower(upper) sums, over the set of all possible partitions, does NOT imply anything about limits of arbitrary sequences of partitions.. Of course there must at least exist such sequences which converge to the correct value, whenever the supremum(infimum) exists.

    Again, be careful: this is Analysis, you can't afford to make intuitive decisions. Consider

    [tex]f(x)=\left\{\begin{array}{cc}1,&\mbox{ if x is rational }\\0, & \mbox{ otherwise }\end{array}\right [/tex].​

    No matter how finely you partition your [0,1] interval, you lower sum is always going to be zero, and your upper sum is always one. The 'error' cannot converge to zero; the Reimann integral does not exist. (edit: It will exist at least when the function is bounded and piecewise continuous.

    edited: I hate LaTeX!!! :grumpy:
    Last edited by a moderator: Jun 16, 2005
  6. Feb 6, 2010 #5
    This may not be true. Think about this: [tex]P_k=\{a, b-\sum\limits_{n=1}^{k-1}\frac{b-a}{2^n}}, b-\sum\limits_{n=1}^{k-2}\frac{b-a}{2^n}}, \cdots, b-\frac{b-a}{4}-\frac{b-a}{2}, b-\frac{b-a}{2}, b\} [/tex]

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