I need some help with something I figured could be proven (which I thought would be easy, but I guess I am wrong [?]) - - I just can't completely see all the reasoning.
I'm working from the following paper that can be found here. See the end of page 76 to the beginning of page 77, in particular. I'll reiterate some of the definitions and assumptions for convenience (where I've only slightly modified the notation from within the literature here, again, for convenience purposes).
We let $(\Omega, \Sigma, \mu)$ be a measure space where $\Omega\neq\varnothing$, and $\mu$ is complete, positive, $\sigma$-finite, and countably additive measure on $\Sigma$. Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$. $L^{p}(\mu)$ is the Banach space of all (equivalence classes of) measurable functions $f:\Omega\rightarrow\mathbb{K}$ for which $\parallel\!~f~\!\!\!\parallel_{p}<+\infty$, $\parallel\!~f~\!\!\!\parallel_{1}:={\displaystyle{\int_{\Omega}|f|d\mu}}$, $\parallel\!~f~\!\!\!\parallel_{+\infty}:=\text{ess-sup}\big\{|f(\omega)|\in\mathbb{K}:\omega\in\Omega\big\}$, and $\parallel\!~f~\!\!\!\parallel_{0}:={\displaystyle{\sum\limits_{n=1}^{+\infty}\dfrac{1}{2^{n}}\dfrac{1}{\mu(E_{n})}\int_{E_{n}}\dfrac{|f|}{1+|f|}d\mu}}$; furthermore, $(E_{n})_{_{n=1}}^{^{+\infty}}$ partitions $\Omega$ into sets such that for all $n\in\mathbb{N}$ we have $0<\mu(E_{n})<+\infty$ - the partition exists since $\mu$ is assumed to be $\sigma$-finite (this I've already shown myself, and is straightforward by definition). Lastly (for the sake of my question) if $\mu$ is finite, we have the following $simpler$ definition $\parallel\!~f~\!\!\!\parallel_{0}:={\displaystyle{\int_{\Omega}\dfrac{|f|}{1+|f|}d\mu}}$.
My Question: I need help with, going by the original definition of $\parallel\!~f~\!\!\!\parallel_{0}:={\displaystyle{\sum\limits_{n=1}^{+\infty}\dfrac{1}{2^{n}}\dfrac{1}{\mu(E_{n})}\int_{E_{n}}\dfrac{|f|}{1+|f|}d\mu}}$, proving, or showing, the $simpler$ definition of $\parallel\!~f~\!\!\!\parallel_{0}:={\displaystyle{\int_{\Omega}\dfrac{|f|}{1+|f|}d\mu}}$ holds whenever we assume $\mu$ is finite (as well as $\sigma$-finite, etc., of course).
I feel this can be accomplished since ${\displaystyle{\sum\limits_{n=1}^{+\infty}\dfrac{1}{2^{n}}=1}}$ as well as $\mu\bigg({\displaystyle{\bigcup\limits_{n=1}^{+\infty}E_{n}\bigg)=\sum\limits_{n=1}^{+\infty}\mu(E_{n}) }}=\mu(\Omega)<+\infty$ since $(E_{n})_{_{n=1}}^{^{+\infty}}$ partitions $\Omega$. Can anyone help me show this??? As always, hints, recommendations, suggestions, answers, etc., are GREATLY appreciated!
A more transparent definition of $\|f\|_0$ would be $$ \|f\|_{0,g}:=\int_\Omega {|f|\over 1+|f|}g\,d\mu, $$ where $g:\Omega\to(0,\infty)$ is measurable and $\int_\Omega g\,d\mu<\infty$. Your specific choice of $g$ is $$ \sum_{n=1}^\infty {1\over 2^n\mu(E_n)}1_{E_n}. $$ Of course, when $\mu(\Omega)<\infty$ the choice $g\equiv 1$ is available.
These $L^0$ "norms" are equivalent, for different choices of $g$, because regardless of the choice of $g$, $\|f_n-f\|_{0,g}\to 0$ if and only if $f_n\to f$ locally in measure (with respect to $\mu$), in the sense that $$ \lim_{n\to\infty}\mu(x\in D:|f_n(x)-f(x)|>\epsilon)=0,\qquad\forall\epsilon>0, $$ for each measurable set $D$ with $\mu(D)<\infty$.