$\sigma$-finiteness of measure I am doing the following problem:
Let $(X,\mathcal{M},\mu)$ be a measure space, $0\leq f<\infty$ is a measurable function, and $d\nu = fd\mu$. Show:
(a) If $\mu$ is $\sigma$-finite, then so is $\nu$.
(b) Assume no longer that $\mu$ is $\sigma$-finite, but assume instead that $0<f<\infty$. Then if $\nu$ is semifinite, then $\mu$ too is.
My attempt for (a):
Assume first that $f$ is integrable. Then $\int_{n}^{n+1}fd\mu=\nu([n,n+1))<\infty$ for any $n\in\mathbb{Z}$. If $f$ is bounded, say, by $M$, then every integral $\int_n^{n+1}fd\mu<M$, still showing that $\nu$ is $\sigma$-finite.
The part where I am stuck is when $f$ is neither bounded nor integrable. We can take $f=1/x^2$, and the partition of $\mathbb{R}$ I chose clearly does not work.
For (b):
Again, if $f$ is integrable, there is no difficulty: If $E$ is a measurable set such that $\nu(E)=\int_Efd\mu>0$, there is $F\subset E$ such that $0<\nu(F) = \int_F fd\mu<\infty$. Clearly $F$ cannot be of measure 0 w/r/t $\mu$, and since $\int_Ffd\mu<\int fd\mu<\infty$, $\mu$ is semifinite.
I am not sure how to proceed from here if $f$ is just bounded, or let alone just measurable (with $0<f<\infty$, of course).
 A: For a) let $A_n:=f^{-1}([n,n+1))$ for $n\in \Bbb N $ and set $f_n:=\mathbf{1}_{A_n}f$. Thus its easy to see that $f_n\,\mathrm d \mu $ is $\sigma $-finite for each $n\in \Bbb N $ and that $f\,\mathrm d \mu =\sum_{n\geqslant 0}f_n\,\mathrm d \mu $, so $\nu $ is also $\sigma $-finite.$\Box$

UPDATE: for b) using the definition of semi-finiteness in the comment, let
$$
B_n:=f^{-1}((n^{-1},n)), \quad n\in \Bbb N_{> 0} \tag1
$$
Now suppose that $\mu (A)=\infty $, and note that $\bigcup_{n>0}(A \cap B_n)=A$. Then by the continuity from below of $\mu $ we have that
$$
\mu (A)=\mu \left(\bigcup_{n>0}(A \cap B_n)\right)=\lim_{n\to \infty }\mu (A \cap B_n)=\infty\tag2 
$$
so there is some $n\in \Bbb N_{> 0} $ such that $\mu (A \cap  B_n)>0$ and from the definitions we have that
$$
\frac1{n}\,\mu (A \cap B_n)\leqslant \nu (A \cap B_n)=\int_{A \cap B_n}f\,\mathrm d \mu \leqslant n\,\mu (A \cap B_n)\tag3
$$
If $\mu (A \cap B_n)=\infty $ then by $\mathrm{(3)}$ we knows that $\nu (A \cap B_n)=\infty $ also, and because $\nu $ is semi-finite then there is some $C\subset (A \cap B_n)$ such that $\nu (C)$ is finite and positive. But by the definition of $B_n$, $\nu $ and the fact that $C\subset B_n$ this immediately imply that $\mu (C)$ is also finite and positive.$\Box$
