I haven't found any other question that answered my problem so…
If $μ,ν $ are two measures in measure space $(X, \mathcal{A})$, $f:X\rightarrow [0,\infty]$ measurable function such that $ν(A)=\int_{A}fdμ$ $\forall A\in\mathcal{A} $ prove that $ν$ is σ-finite iff $f<\infty$ $μ$– almost everywhere.
I have proved$(\Rightarrow) $. For $ (\Leftarrow)$ I only thought this:
for $Z=[f=\infty]$(elements in $X$ for which $f(x)=\infty$), $μ(Ζ)=0<\infty$ so $ν(Ζ)<\infty$. Let $B=X\backslash Z$ and $A_n=\{x \in B: n\leq f<n+1\}\in \mathcal{A}$ because $ f$ is measurable.
I tried to prove that $ν(A_n)<\infty$ but I don't think it's true. Can someone help me?
Best Answer
This is clearly false without some extra assumption on $\mu$. If $f \equiv 1$ then $\nu =\mu$ so $\nu$ cannot be $\sigma$-finite unless $\mu$ is.
However, if $\mu$ is $\sigma$-finite then the assertion is true. Suppose $B_n$'s are disjoint sets of finite $\mu$-measure whose union is $X$. Let $A_n$'s and $Z$ be as in your attempt. Then $\nu (A_n \cap B_m) \leq (n+1) \mu (B_n) <\infty$ for all $n,m$ and $X$ is covered by the countable collection of sets $\{Z, A_n \cap B_m : n,m \geq 1\}$.