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Integration with respect to the counting measure

  1. Jan 14, 2012 #1
    I am struggling with convincing myself that if you equip [itex]\mathbb Z[/itex] with the counting measure [itex]m[/itex], the [itex]L^p[/itex] norm of measurable functions [itex]f: \mathbb Z \to \mathbb C[/itex] looks like
    \| f \|_p = \left( \sum_{n = -\infty}^\infty |a_n|^p \right)^{1/p}.
    I know that any function on [itex]\mathbb Z[/itex] is essentially just a doubly-infinite sequence of complex numbers [itex]\left\{ \ldots, a_{-2}, a_{-1}, a_0, a_1, a_2, \ldots \right\}[/itex], with [itex]a_n = f(n)[/itex]. But how does one get from the general definition of integrals of positive functions (which [itex]|f|^p[/itex] certainly is) to the sum that appears above?
  2. jcsd
  3. Jan 14, 2012 #2
    So you wish to prove that (if c is the counting measure)

    [tex]\int fdc=\sum_{-\infty}^{+\infty}f(n)[/tex]

    Let's do this in steps:

    1) f is of the form [itex]I_{\{x\}}[/tex] (where [itex]I_A(z)=1[/itex] if [itex] z\in A[/itex] and 0 otherwise), then

    [tex]\int f dc=\int I_{\{x\}}dc=c\{x\}=1[/tex].

    2) f is of the form [itex]\sum_{k=-n}^n{a_k I_{\{x_k\}}}[/itex] (a finite combination of things in (1)). Then

    [tex]\int fdc=\sum_{k=-n}^n a_k \int I_{\{x_k\}} = \sum_{k=-n}^n{a_k}[/tex]

    3) If f is positive of the form [itex]\sum_{k=-\infty}^{+\infty} {a_k I_{\{x_k\}}}[/itex] (with [itex]a_k\geq 0[/itex]). Then we apply the monotone convergence theorem:

    [tex]\int fdc= \sum_{k=-\infty}^{+\infty}{a_k}[/tex]

    4) Finally, if f is integrable, then we can write [itex]f=f^+-f^-[/itex] and thus

    [tex]\int fdc=\int f^+dc-\int f^-dc[/tex]
  4. Jan 14, 2012 #3
    Yeah. Thinking about it using the monotone convergence theorem the way you did definitely seems like the best approach. Thanks, micromass.

    By the way...if your measure space is [itex](\Omega, \mathcal P(\Omega))[/itex] (a set and its power set), isn't every function [itex]f: \Omega \to Y[/itex], where [itex]Y[/itex] is some topological space, measurable?
  5. Jan 14, 2012 #4
    Yes, if your sigma-algebra is [itex]\mathcal{P}(Y)[/itex], then everything is measurable.
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