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Let $\mu$ be a sigma finite positive measure on $(X,\mathcal{A})$. then exists $w\in L^1(\mu)$ such that $0< w(x) < 1$ for all $x\in X$.

Since $\mu$ is a sigma finite measure we have that $$X=\bigcup_{n=1}^\infty E_n\quad \mu(E_n)<\infty.$$ We define $$w_n(x)=\frac{1}{2^n(1+\mu(E_n))}\quad\text{if}\;x\in E_n$$ zero otherwise.

Define $$w(x):=\sum_{n=1}^\infty w_n(x).$$

I can't find a way to show that $$\int_X w\;d\mu <\infty$$ could someone give me a suggestion?

Why $0<w<1$?

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  • $\begingroup$ I think your definition of $w_n$ is missing a term like $\chi_{E_n}$. $\endgroup$
    – Umberto P.
    Jul 11, 2022 at 17:51

2 Answers 2

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Since our measure is $\sigma$-finite, we may write $X = \bigcup_{n=1}^\infty E_n$, where $\mu(E_n) < \infty$ And $E_n \subseteq E_{n+1}$ for all $n$. For a set $A$, let $I_A(x)$ denote the indicator function (also called characteristic function) for the set $A$. Now set: $$\begin{align*} w_n(x) & := \frac{I_{E_n}(x)}{2^{n}(1 + \mu(E_n))} \quad n \in \mathbb{N} \\ S_n(x) & := \sum_{j=1}^n w_j(x) \\ w(x) & := \sum_{j=1}^\infty w_j(x) \end{align*} $$ We may compute: $$\begin{align*} \int_X S_n \: d\mu & = \sum_{j=1}^n \int_{X} \omega_n \: d\mu \\ & = \sum_{j=1}^n \frac{1}{2^{n}(1 + \mu(E_n))} \int_X I_{E_n} \: d \mu \\ & = \sum_{j=1}^n \frac{\mu(E)}{2^{n}(1 + \mu(E_n))} \\ & < \sum_{j=1}^n \frac{1}{2^n} \\ & \leq 1 \end{align*} $$ Further, we see that $S_n \nearrow \omega$. hence by the monotone convergence theorem, we have that $\int_X \omega \: d \mu \leq 1 < \infty$.

We also see that for any $x \in X$, there is some $E_k$ such that $x \in E_k$. So we compute: $$ \begin{align*} \omega(x) & \geq \omega_k(x) \\ & = \frac{1}{2^k(1 + \mu(E_k))} \\ & > 0 \end{align*}$$

We also have that: $$ \begin{align*} \omega(x) & = \sum_{n=1}^\infty \frac{I_{E_n}(x)}{2^n(1 + \mu(E_n))} \\ & \leq \sum_{n=1}^\infty \frac{1}{2^n(1 + \mu(E_n))} \\ & < \sum_{n=1}^\infty \frac{1}{2^n} \\ & = 1 \end{align*}$$ So $0 < w < 1$.

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  • $\begingroup$ You need to get the strict inequality $w<1$, which I know is easy. $\endgroup$
    – Mateo
    Jul 11, 2022 at 19:29
  • $\begingroup$ Sorry! Should be complete now. $\endgroup$
    – Joe
    Jul 11, 2022 at 20:27
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Actually, you can modify your construction by choosing the sequence $(E_n)$ to be pairwise disjoint instead of increasing. In this way, $$ \int_X w(x)d\mu(x)=\sum_{n\geqslant 1}\int_X w_n(x)d\mu(x)\leqslant \sum_{n\geqslant 1}2^{-n} $$ and since for each $x$, there exists exactly one $n(x)$ such that $x\in E_{n(x)}$, we have $0<w(x)\leqslant 2^{-n(x)}<1$.

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