# A particular function of $L^1$

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?

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

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 $$0 < w < 1$$.
• You need to get the strict inequality $w<1$, which I know is easy. Jul 11, 2022 at 19:29
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.