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Let ${(X_n, \mathcal A_n, \mu_n)_{n \in \mathbb N}}$ be measure spaces. Let $X_n$ be pairwise disjoint. Define $(X, \mathcal A, \mu)$ by$X=\bigcup_{i=1}^{\infty} X_n$

$\mathcal A=\{E \subset X: E \cap X_n \in \mathcal A_n \; \forall n \in \mathbb N\}$ and $\mu(E)=\sum_n \mu_n(E \cap X_n)$

I want to show that if all $\mu_n$ are $\sigma$-finite than $\mu$ is $\sigma$-finite aswell.

My attempt: I wanted to take the sequence of the unions of the sequences that make $\mu_n$ sigma finite. So the first element is the union of all the first parts of the each sequence and so on.

Showing that the union is of all these equals $X$ is easy, but I fail at showing that each element has finite measure and that each element is contained in $A$ and hoped that someone could help me here.

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  • $\begingroup$ Hint: Do you remember how to prove that $\mathbb N \times \mathbb N$ is countable? $\endgroup$ – Kenny Wong Apr 27 '17 at 17:39
  • $\begingroup$ @KennyWong You mean first listing all tupels, who start with $1$ then all starting with $2$ etc. ? I can see that there is a connection in the idea but not how to explicitly use it $\endgroup$ – PeterGarder Apr 27 '17 at 18:46
  • $\begingroup$ But that doesn't work. Since there are infinitely many tuples starting with $1$, you'll never reach the ones starting with $2$. You have to enumerate the tuples "diagonally". It's hard to explain without drawing a picture, so perhaps you could look at this: homeschoolmath.net/teaching/rational-numbers-countable.php Anyway, the problem you posed is similar: it boils down to showing that a countable union of countable collections of "things" is itself a countable union of "things". In your case, the "things" are the measurable sets with finite measure. $\endgroup$ – Kenny Wong Apr 27 '17 at 19:23
  • $\begingroup$ What are $E$ and $B$ exactly in your definition of $\mathcal{A}$ and $\mu(E)$? Can you fix those definitions? $\endgroup$ – gogurt Apr 27 '17 at 19:29
  • $\begingroup$ @gogurt I am sorry, $B$ was meant to be $E$. It's edited now. $E$ in $\mu(E)$ is an element of $\mathcal A$ $\endgroup$ – PeterGarder Apr 27 '17 at 19:33
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As the comments indicate, you need to accept that countable unions of countable sequences are still countable. See here for example.

Now you just need to show that the measure $\mu$ is $\sigma$-finite, i.e. you need to show that $X$ is a countable union of sets all of which have finite measure. As you said, do this using the sequences which make each $\mu_n$ $\sigma$-finite.

Let $\{U_{i,k}: k \in \mathbb{N}\}$ be a countable sequence of sets in $\mathcal{A}_i$ which satisfy

  • $\mu_i(U_{i,j}) < \infty$ for all $j \in \mathbb{N}$
  • $\bigcup_j U_{i,j} = X_i$

We will consider collection $\{U_{i,j}, i \in \mathbb{N}, j \in \mathbb{N}\}$. All we need to show is that this collection 1) is countable, 2) unions up to $X$, and 3) all elements have finite measure under $\mu$.

Using the fact above, 1 is immediate. 2 is also obvious. 3 is just a tiny bit of work. Pick an arbitrary $U_{i,j}$ from that collection. This set is in $\mathcal{A}$ because

  1. $U_{i,j} \subset X$
  2. $U_{i,j} \cap X_i = U_{i,j}$ which is in $\mathcal{A}_i$, and
  3. $U_{i,j} \cap X_k$ for $i \neq k = \emptyset$ since the $X_k$'s are pairwise disjoint, and $\emptyset \in \mathcal{A}_k$ for any $k$.

We already know that $\mu_i(U_{i,j}) < \infty$ and obviously $\mu_j(\emptyset) = 0$ for any $j \neq i$ so that $\mu(U_{i,j}) < \infty$ also.

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