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How I can show that every finite measure can be regarded as a $\sigma$-finite measure but not conversely in general?

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Please write down definition if sigma-finitness. Think of usage of empty sets – Norbert May 13 '12 at 15:00
For the converse, consider Lebesgue measure on $\Bbb R$. – David Mitra May 13 '12 at 15:55

Let $(X,\mathcal A,\mu)$ a measure space, where $\mu$ is a non-negative measure. $\mu$ is said finite if $\mu(X)<\infty$. The space $(X,\mathcal A,\mu)$ is $\sigma$-finite if we can find a sequence $\{A_n\}_{n=1}^{+\infty}$ of measurable sets of finite measure such that $X=\bigcup_{n\geq 1}A_n$.

  • If $(X,\mathcal A,\mu)$ is finite, it $\sigma$-finite, since you can take $A_1:=X$ and $A_j=\emptyset$ for $j\geq 2$.
  • But the converse is not true. Take $X$ an infinite countable set, $\mathcal A:=2^X$ the collection of all the subsets of $X$ and $\mu$ the counting measure ($\mu(A)$ is the number of elements of $A$ if $A$ is finite and $+\infty$ otherwise). It's $\sigma$-finite space, because if $X=\{x_n,n\in\Bbb N\}$, we can write $X=\bigcup_{n\geq 1}\{x_{n}\}$ and $\forall n\in \mathbb{N}:$ $\mu(\{x_{n}\})=1<\infty$, but not finite since $X$ is infinite.
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I'm going to add another counter-example on top of Davide's one, which is also quite simple. As Davide pointed out a finite measure is $\sigma$-finite trivially by choosing rest of the sequence as empty sets.

Consider the Lebesgue measure $m_{n}$ on $\mathbb{R}^{n}$. Now $\mathbb{R}^{n}=\bigcup_{k=1}^{\infty}B(\bar{0},k)$ and each $B(\bar{0},k)$ has finite measure (e.g. by observing that $m_{n}(B(\bar{0},k))\leq (2k)^{n}$ for all $k\in\mathbb{N}$). Yet $m_{n}(\mathbb{R}^{n})=\infty$.

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