I want to prove the following:
Theorem: Let $(\Omega, \mathcal{A}, \mu)$ be a measure space. Then $\Omega$ is $\sigma$-finite if and only if there exists an $f\colon \Omega \longrightarrow (0,\infty)$ and $f\in\mathcal{L}^1(\mu)$.
Proof:
Suppose the measure space is $\sigma$-finite, then there exists a sequence $(\Omega_n)_{n\in\mathbb{N}}\in \mathcal{A}$ with $\Omega=\bigcup_{n=1}^\infty \Omega_n$ and $\mu(\Omega_n)<\infty,\ \forall n\in\mathbb{N}$.
I need to construct a function as stated in the theorem, wich is integrable. Since, we are given the sequence $(\Omega_n)_{n\in\mathbb{N}}$ it is reasonable to try a simple function like
$$f:=\sum_{n=1}^\infty \frac{1}{\mu(\Omega_n)2^n}\chi_{\Omega_n}$$ where $\chi$ is the indicator function. By construction $f$ is positive and
$$\int_\Omega |f|\,\mathrm{d}\mu=\lim_{N\rightarrow\infty}\sum_{n=1}^N\frac{\mu(\Omega_n)}{\mu(\Omega_n)}\frac{1}{2^n}\chi_{\Omega_n}=1$$
In the second equality I used the monotone convergence theorem. This proves one direction.
For the other direction I use that $f$ has an expansion as a limit of simple functions $f=\sum_{n=1}^\infty \alpha_i\chi_{A_i}, \alpha_i>0, A_i\in\mathcal{A}, \ \forall i\in\mathbb{N}$. Again, by the monotonce convergence theorem:
$$\int_X |f|\,\mathrm{d}\mu=\lim_{n\rightarrow\infty}\sum_{i=1}^n\alpha_i\mu(A_i)<\infty$$
It is important, that $f$ is strictly positive, because it might happen, that $\mu(A_i)=\infty$ for some $i$, and then $\alpha_i\mu(A_i)=\infty$ which will not allow me to conclude. Instead, we always have $\alpha_i\mu(A_i)<\infty$ from which it follows that $\mu(A_i)<\infty$. (Note that $0\cdot\infty:=0$) Finally, note that $X=\bigcup_{i=1}^\infty A_i$ and we are done.
Is this proof correct? Also I am wondering if the proof can be extended for functions $f\colon \Omega \longrightarrow (0,\infty]$, that is including infinity.
Best Answer
Your idea is correct but there are notational problems. Each approximating simple function is a finite sum $ \sum a_i I_{A_i}$ but the numbers $a_i$ and sets $A_i$ themselves vary with $n$. You need a double indexed family $A_{i,n}$ and then looking at the union of all these sets you can finish the proof.
If $f$ takes values in $(0,\infty]$ and $\int f d\mu <\infty$ then automatically $f<\infty$ almost everywhere so the result does extend to this case.
I would suggest an alternative simpler proof: let $A_n=\{x: f(x) >\frac 1 n\}$. Then $\int f d\mu \geq \int_{A_n} fd\mu \geq \frac 1 n \mu (A_n)$ so $\mu (A_n) <\infty$ for each $n$. Also $\bigcup_n A_n =\Omega$.