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Let $(X,\mathcal{A}, \mu)$ be a measure space. Prove:

$\mu$ is $\sigma$-finite $\iff \exists$ function $f \in \mathcal{L}^{1}(\mu)$ with $f(x)>0$ for all $x \in X$

My ideas

$"\Leftarrow"$

Let $f \in \mathcal{L}^{1}(\mu)$ with $f(x)>0$, $\forall x \in X$. So, $f$ is measurable. This implies that for a $B_{n}$ defined as $B_{n}:=\{f>\frac{1}{n}\}$ which is measurable, so $\in \mathcal{A}$. By definition, $\{f>0\}=\bigcup_{n\in \mathbb N}B_{n}\in \mathcal{A}$, but since $f > 0, \forall x \in X$ then $X\subseteq\bigcup_{n\in \mathbb N}B_{n}$ and $\mu(B_{n})<\infty, \forall n \in \mathbb N$, since $\int_{X}fd\mu < \infty$$\Rightarrow \mu$ is $\sigma-$finite.

$"\Rightarrow"$ I have no idea how to define this function, particularly as $f>0$, $\forall x \in X$

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1 Answer 1

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Let $(E_n)_{n \in \mathbb{N}}$ be measurable (disjoint) sets with $\mu(E_n) < \infty$ and $\Omega = \bigcup_{n=1}^\infty E_n$. Now define $$f(x) := \sum_{n=1}^\infty \frac{1}{2^n} \frac{1}{1 + \mu(E_n)} 1_{E_n}(x).$$ By definition, we have $f(x) >0$ for all $x \in \Omega$. On the other hand, we have $$\int f(x) d \mu(x) \le \sum_{n=1}^\infty 2^{-n} \frac{\mu(E_n)}{1+\mu(E_n)} \le 1. $$

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