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Let $\mu$ be a $\sigma$-finite measure on $(A,\mathcal{A})$. Then there are finite measures $(\mu_n)_{n \in \mathbb{N}}$ on $(X,\mathcal{A})$ such that $$\mu = \sum_{n \in \mathbb{N}}\mu_n$$

So if $\mu$ is $\sigma$-finite, we have that $$X = \bigcup_{n \in \mathbb{N}}X_n$$ for some measurable sets $X_n$ with $\mu(X_n) < \infty$ for any $n \in \mathbb{N}$. My first idea was something like restricting this measures to the sets $X_n$ but $\mu_n$ must be defined on $\mathcal{A}$. Any hint?

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You can take the $X_{n}$ disjoint. Define $\mu_{n}(A)=\mu(A\cap X_{n})$ for $A\in \mathcal{A}$.

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  • $\begingroup$ I accept this answer since I think it is necessary that one takes the $X_n$ to be disjoint. $\endgroup$ – TheGeekGreek Jan 26 '17 at 21:17
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Restricting to $X_n$ is the right idea, and the way to accomplish that is to set $\mu_n(A) = \mu(A \cap X_n)$. This makes sense for any $A \in \mathcal{A}$.

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