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This is question 13.2 in "Measures, Integrals and Martingales" by R.L. Schilling. The answer can be found here. I don't understand why they use an exhausting sequence rather than apply the definition of the product measure directly.

Let $(X,\mathcal{A},\mu)$, $(Y,\mathcal{B},\nu)$ be two $\sigma$-finite measure spaces. Show that $A \times N$, where $A \in \mathcal{A}$ and $N \in \mathcal{B}$, $\nu(N) = 0$, is a $\mu \times \nu$ - null set.

Now in the answer, they take two exhausting sequences and by taking the intersection of $A$ and $N$ with these sequences the proposition is proven. I would understand this if the product measure was defined only on those sequences, but it is defined as (theorem 13.5):

$\rho: \mathcal{A} \times \mathcal{B} \to [0,\infty]$, $\rho(A \times B) := \mu(A)\nu(B)$.

Why can't we just say $\rho(A \times N) = \mu(A)\nu(N) = \mu(A) 0 = 0$ ?

Eugene Zhang
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Scipio
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    If "we just say that" then we are also saying that $\infty.0=0$. Care is needed. Maybe you should check what is said about product measures uptil that stage of the book. – drhab Nov 04 '15 at 20:07
  • Thank you! Yes I was thinking that might be the problem. However, it is explicitly said in a previous chapter (ch 8, on the extended real line) that $0*(\infty) = 0$. This is not an expression that we safely use in general then? Chapter 13 is the first chapter on product spaces btw. Sorry if this is a bit elementary. – Scipio Nov 04 '15 at 20:13
  • I have no example at hand where the equality "goes wrong" and am not eager to explore the book. Maybe someone else will help you with this. Good luck! – drhab Nov 04 '15 at 20:20
  • you're right, I can google that much ;) this question gives some insight. Not really used to working with infinity yet... Let me know if this question is useless and I should remove it. – Scipio Nov 04 '15 at 20:27
  • For $A$ is of infinity measure, use the fact that $A$ is $\sigma$-finite and find a sequence of set with finite measure to approach it. – Eugene Zhang Nov 05 '15 at 02:44

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If $\mu(A)<\infty$, you are right. It needs to prove the case for $\mu(A)=\infty$.

Since $\mu$ is $σ$-finite, there is a sequence of sets $O_n$ with finite measure, i.e. $\mu(O_n)<\infty$ such that $$ O_1\subset O_2\subset \cdots\subset O_n\subset \cdots\quad\text{and }\quad A=\bigcup_{n=1}^{\infty} O_n $$ By monotone class theorem in measure theory $$ \mu (A)=\mu \left ( \bigcup _{n=1}^{\infty} O_n\right )=\lim _{n\to \infty }\mu (O_n) $$ So we have $$ \rho(A \times N) = \rho\left ( \bigcup _{n=1}^{\infty} O_n\times N\right ) = \lim _{n\to \infty }\rho (O_n\times N)=\lim _{n\to \infty }\mu (O_n)\nu(N) = 0 $$

Eugene Zhang
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