Measure is sigma-finite

Question

is it true that $(X, \mathbb A, \mu)$ is $\sigma$-finite if and only if there exists a measurable function $f : (X, \mathbb A) \to (\mathbb R, B(\mathbb R))$ which is (strictly) positive, piecewise constant and with finite integral with respect to $\mu$ ?

Answer 1

Yes, the statement is true. Assume that $\mu (X)>0$ and $\mu$ is $\sigma$ finite. Then there exists a sequence of pairwise disjoint sets $A_n \in \mathbb{A}$ such that $0<\mu (A_n ) <\infty$ and $$X=\bigcup_n A_n .$$ Define a function $f:X\to\mathbb{R},$ by $f(t) =2^{-n} (\mu (A_n )^{-1} $ for $t\in A_n .$ Then function $f$ is positive, piecewise constant , Borel measurable with finite integral over $X.$

On the other hand if such a function exists then obviously $\mu $ is sigma finite.

Answer 2

You are asking if it's true that $$\mu \,\,\,is\,\,\, \sigma-finite \iff \exists f=\sum_{k=1}^{n}a_{k}\chi_{A_k} \,\,\, s.t. \,\,\, \int_{X}fd\mu <\infty$$

Well if we recall the definitions of $\sigma$-finite and integral of a simple function, then we are in a good position.

1) $\sigma$-finite: A measure $\mu$ is $\sigma$-finite is there exists sets $A_1 ,A_2 ,...$ such that $X=\cup A_n$ and $\mu(A_n )<\infty \forall n$

2) Integral of a simple function: Given a simple function with canonical representation $f=\sum_{k=1}^{n}a_{k}\chi_{A_k}$, the integral of $f$ over $X$ with respect to $\mu$ is given by $\int_{X}fd\mu = \sum_{k=1}^{n}a_{k}\mu(X\cap A_k)$

Now you can work with the definitions to see the equivalence