5
$\begingroup$

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 $f>0$, $\forall x \in X$

$\endgroup$
9
$\begingroup$

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 $$\int f(x) d \mu(x) \le \sum_{n=1}^\infty 2^{-n} \frac{\mu(E_n)}{1+\mu(E_n)} \le 1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.