Integration with respect to the counting measure

[AxiomOfChoice]

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

 

[micromass]

So you wish to prove that (if c is the counting measure)

\int fdc=\sum_{-\infty}^{+\infty}f(n)

Let's do this in steps:

1) f is of the form I_{\{x\}}[/tex] (where I_A(z)=1 if z\in A and 0 otherwise), then<br /> <br /> \int f dc=\int I_{\{x\}}dc=c\{x\}=1.<br /> <br /> 2) f is of the form \sum_{k=-n}^n{a_k I_{\{x_k\}}} (a finite combination of things in (1)). Then<br /> <br /> \int fdc=\sum_{k=-n}^n a_k \int I_{\{x_k\}} = \sum_{k=-n}^n{a_k}<br /> <br /> 3) If f is positive of the form \sum_{k=-\infty}^{+\infty} {a_k I_{\{x_k\}}} (with a_k\geq 0). Then we apply the monotone convergence theorem:<br /> <br /> \int fdc= \sum_{k=-\infty}^{+\infty}{a_k}<br /> <br /> 4) Finally, if f is integrable, then we can write f=f^+-f^- and thus<br /> <br /> \int fdc=\int f^+dc-\int f^-dc

 

[AxiomOfChoice]

So you wish to prove that (if c is the counting measure)

\int fdc=\sum_{-\infty}^{+\infty}f(n)

Let's do this in steps:

1) f is of the form I_{\{x\}}[/tex] (where I_A(z)=1 if z\in A and 0 otherwise), then<br /> <br /> \int f dc=\int I_{\{x\}}dc=c\{x\}=1.<br /> <br /> 2) f is of the form \sum_{k=-n}^n{a_k I_{\{x_k\}}} (a finite combination of things in (1)). Then<br /> <br /> \int fdc=\sum_{k=-n}^n a_k \int I_{\{x_k\}} = \sum_{k=-n}^n{a_k}<br /> <br /> 3) If f is positive of the form \sum_{k=-\infty}^{+\infty} {a_k I_{\{x_k\}}} (with a_k\geq 0). Then we apply the monotone convergence theorem:<br /> <br /> \int fdc= \sum_{k=-\infty}^{+\infty}{a_k}<br /> <br /> 4) Finally, if f is integrable, then we can write f=f^+-f^- and thus<br /> <br /> \int fdc=\int f^+dc-\int f^-dc

<br /> Yeah. Thinking about it using the monotone convergence theorem the way you did definitely seems like the best approach. Thanks, micromass.<br /> <br /> By the way...if your measure space is (\Omega, \mathcal P(\Omega)) (a set and its power set), isn&#039;t <b>every</b> function f: \Omega \to Y, where Y is some topological space, measurable?

 

[micromass]

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 (\Omega, \mathcal P(\Omega)) (a set and its power set), isn't every function f: \Omega \to Y, where Y is some topological space, measurable?

Yes, if your sigma-algebra is \mathcal{P}(Y), then everything is measurable.