# Integration with respect to the counting measure

1. Jan 14, 2012

### 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
$$\| f \|_p = \left( \sum_{n = -\infty}^\infty |a_n|^p \right)^{1/p}.$$
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?

2. Jan 14, 2012

### 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 $I_{\{x\}}[/tex] (where [itex]I_A(z)=1$ if $z\in A$ and 0 otherwise), then

$$\int f dc=\int I_{\{x\}}dc=c\{x\}=1$$.

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

$$\int fdc=\sum_{k=-n}^n a_k \int I_{\{x_k\}} = \sum_{k=-n}^n{a_k}$$

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:

$$\int fdc= \sum_{k=-\infty}^{+\infty}{a_k}$$

4) Finally, if f is integrable, then we can write $f=f^+-f^-$ and thus

$$\int fdc=\int f^+dc-\int f^-dc$$

3. Jan 14, 2012

### AxiomOfChoice

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?

4. Jan 14, 2012

### micromass

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

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