Prove if $f$ and $g$ are integrable, $\int g\,dv=\int fg \, d\mu$

Question

Suppose $(X,A,\mu)$ is a measure space and $f\ge0$ is integrable. $$v(A)=\int_Af \, d\mu$$

1. $v$ is a measure.
2. If $g$ is also integrable with respect to $v$, then $fg$ is integrable with respect to $\mu$ and $$\int g\,dv=\int fg \, d\mu$$

I got stuck with proving the sigma additivity for 1) and 2). Anything would be helpful!

Answer 1

Hint

To prove your equality, prove it first for $g=1_A$ for $A$ measurable, then for $g$ a simple function, and to conclude use the density of simple function in $L^1$.

Edit

1) Let $g=1_A$ where $A$ is measurable. Then, $$\int gd\nu=\int_Ad\nu=\nu(A)=\int_A fd\mu=\int gfd\mu.$$

2) Let $g$ simple, i.e. $g=\sum_{k=1}^na_i 1_{A_i}$ where $A_i$ are measurable.

$$\int gd\nu=\sum_{i=1}^n\int_{A_i}a_id\nu=\sum_{i=1}^na_i\nu(A_i)=\sum_{i=1}^n\int_{A_i} a_ifd\mu=\int \sum_{i=1}^n a_i1_{A_i}fd\mu=\int gfd\mu.$$

3) Let $g\geq 0$ measurable. There is a sequence of simple function s.t. $\varphi_n\nearrow g$ (i.e. $\varphi_n\to g$ and $\varphi_n$ increasing). Using monotone convergence theorem twice, $$\int gd\nu=\lim_{n\to \infty }\int \varphi_nd\nu=\lim_{n\to \infty }\int \varphi_n fd\mu=\int gfd\mu.$$

4) If $g$ measurable, write $g=g^+-g^-$ where $g^+(x)=\max\{0,g(x)\}$ and $g^-(x)=-\min\{g(x),0\}$. Remark that $g^+,g^-\geq 0$. Then, $$\int gd\nu=\int g^+d\nu-\int g^- d\nu=\int g^+fd\mu-\int g^- fd\mu=\int(g^+-g^-)fd\mu=\int fg d\mu.$$