Prove that if $\nu(A)=\int_A fd\mu$ and $\mu$ is $\sigma$-finite, then $\nu$ is $\sigma$-finite. Let $\nu$ be the measure defined on a $\sigma$ finite measure space with measure $\mu$ and a non negative, measurable, real valued function $f$ such that $\nu=f*\mu$ so that $\nu(A)=\int_Afd\mu$.I know that $\nu$ is a measure. My attempt has been to construct a nested sequence of measurable sets $A_i$ so that $A_i\subset A_{i+1}$ and $\Sigma_{i\in \mathbb{N}}=\Omega$. Now because $\mu$ is $\sigma$- measurable $\mu(A_i)<\infty$ $\forall A_i$. Now if $\nu(A_N)=\infty$ it follows from the linearity of the integral that $\nu(A_{i>N})=\infty$. However I haven't been able to produce a contradiction from this assumption.
 A: What you have written is in general wrong with the current assumptions. Consider the measure space $(\mathbb{R}.\mathcal{B},\lambda)$ the standard measure space on the reals with the Lebesgue measure. Then consider
$$f=\begin{cases} \frac{1}{x} & \text{for }  x>0 \\ 
0 & \text{for } x \leq0 
\end{cases},$$
which satisfies your assumptions on $f$.
Then by monotonicity of the integral $$\nu((-1,1))=\int_0^1 fd\lambda \geq \sum_{n=1}^\infty 2^n \int_{2^{-n-1}}^{2^{-n}}d\lambda = \frac12+\frac12+\frac12+\dots = \infty,$$ and due to the nesting propert of $\sigma$-finite measures there exists a large enough $N$ such that $(-1,1) \subset A_N$ and by monotonicity of measures $\nu(A_n)=\infty \quad \forall n\geq N$,
hence $\nu$ can't possibly be sigma finite.
A: I could be wrong, but the following was what I had in mind when I suggested to approximate $f$ with simple functions.
Let $E_n$ be the sequences of sets through which $\mu$ is $\sigma$ finite.
If $f=\chi_A$, then $v(E_n)<\infty $. Now if $v_1$ and $v_2$ are $\sigma$ finite measures (with sequences $I_n$ and $J_m$), then $v_1+v_2$ is a $\sigma$ finite measure as well:
$$(v_1+v_2)(I_n\cap J_m)<\infty$$
and $$\bigcup_{n=1}^{\infty}\bigcup_{m=1}^{\infty} I_n \cap J_n = X$$
It doesn't take much to conclude that linear combinations of $\sigma$ finite measures is also a $\sigma$ finite measure. Let $s_n$ be a simple function such that $|\int f - s_n|<\epsilon$,  then $$\int f <\int s_n +\epsilon$$
Notice that, $\int s_n = s_n * \mu$ and $s_n$ is a linear combination of characteristic functions. So $s_n * \mu$ is a $\sigma$ finite measure for some sequence of sets $F_n$. As a result, $\int f$ is a sigma finite measure for some sequence of sets $F_n$.
