Finite measures are $\sigma$-finite

How I can show that every limited action can be considered as a $\sigma$ - limited action yet not alternatively as a whole?

9
2022-07-25 20:43:41
Source Share
Answers: 1

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.
13
2022-07-25 21:13:05
Source