Let $(X,B, \mu)$ be a complete measure and $(\mathbb{N},P(\mathbb{N}),\nu)$ be counting measure.
Are the following true?
a) $A \subset X \times \mathbb{N}$ if and only if $A^n \in B$ for all $n \in \mathbb{N}$.
b)$f(x,n)$ is $\mu \times \nu$– measurable if and only if $f^n(x)=f(x,n)$ is $\mu$-measurable for all $n\in \mathbb{N}$.
c)If $f\in L^+(\mu \times \nu)$, then $\int f d\mu \times \nu = \sum_{n \geq 1} \int f^n(x) d\mu(x)$, where $\mu$ is not necessarily $\sigma$-finite.
My attempt:
a) true, by Proposition 2.34 in Folland.
b) true, by Proposition 2.34 in Folland.
c) false, since we need $\sigma$– finiteness in Fubini-Tonelli.
I'm not sure if the above is correct, please let me know if something is wrong… Thank you!
Best Answer
(a) Consider measurable spaces $(X,\mathcal{F})$ and $(\mathbb{N},\mathcal{P}(\mathbb{N}))$. Let $(X\times\mathbb{N},\mathcal{F}\otimes\mathcal{P}(\mathbb{N}))$ be their product. Let $A\subseteq X\times N$. We go to prove that $A\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$ iff for each $n\in\mathbb{N}$, $A^{n}\in\mathcal{F}$, where $A^{n}=\{x\in X\mid(x,n)\in A\}$.
Proof: Let $\pi_{X}:X\times\mathbb{N}\rightarrow X$ and $\pi_{\mathbb{N}}:X\times\mathbb{N}\rightarrow\mathbb{N}$ be the canonical projections, i.e., $\pi_{X}(x,n)=x$ and $\pi_{\mathbb{N}}(x,n)=n$. For each $x\in X$ and $n\in\mathbb{N}$, define $i_{x}:\mathbb{N}\rightarrow X\times\mathbb{N}$ by $i_{x}(n)=(x,n)$ and $j_{n}:X\rightarrow X\times\mathbb{N}$ by $j_{n}(x)=(x,n)$. Since the product $\sigma$-algebra $\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$ is the smallest $\sigma$-algebra on $X\times\mathbb{N}$ such that $\pi_{X}$ and $\pi_{\mathbb{N}}$ are measurable, it can be proved easily that $\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$ has the following universal property:
For any measurable space $(Y,\mathcal{M})$ and map $g:Y\rightarrow X\times\mathbb{N}$, $g$ is $\mathcal{M}/\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$-measurable iff $\pi_{X}\circ g$ and $\pi_{\mathbb{N}}\circ g$ are measurable.
Now $\pi_{X}\circ j_{n}=id_{X}$ (the identity map on $X$) and $\pi_{\mathbb{N}}\circ j_{n}:X\rightarrow\mathbb{N}$ is the constant function $x\mapsto n$, which are both measurable. Therefore, $j_{n}$ is measurable. Similarly, $i_{x}$ is also measurable.
$\Rightarrow:$ Let $A\subseteq X\times\mathbb{N}$. Suppose that $A\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$. Let $n\in\mathbb{N}$ be arbitrary. Note that $A^{n}=j_{n}^{-1}(A)$ and hence $A^{n}\in\mathcal{F}$ because $j_{n}$ is measurable.
$\Leftarrow:$ Let $A\subseteq X\times\mathbb{N}$. Suppose that for each $n\in\mathbb{N}$, $A^{n}\in\mathcal{F}$. For each $n\in\mathbb{N}$, $A^{n}\times\mathbb{N}=\pi_{X}^{-1}(A^{n})\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N}).$ Observe that $X\times\{n\}=\pi_{\mathbb{N}}^{-1}(\{n\})\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$. Therefore, $\left(A^{n}\times\mathbb{N}\right)\cap\left(X\times\{n\}\right)\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$. Finally, $A=\cup_{n}\left(A^{n}\times\mathbb{N}\right)\cap\left(X\times\{n\}\right)\in\mathcal{F}\otimes\mathcal{P}(\mathbb{N})$.
Remark: In the above, if we replace $(\mathbb{N},\mathcal{P}(\mathbb{N}))$ by an arbitrary measurable space $(Y,\mathcal{G})$, $\Rightarrow$ still holds. However, in general, $\Leftarrow$ fails to hold. (Notice that, in the proof of $\Leftarrow$, we use the fact that $\mathbb{N}$ is countable in the last step.)