Ytukyg

Let $mu$ be a measure on $(X, mathcal{A})$ and a measurable function $f:X to mathbb{R}, f geq 0$.

Define $mu_f(E): mathcal{A} to mathbb{R}, mu_f(E):=int_E f dmu$ for $E in mathcal{A}$.

How to prove that $mu_f$ is a measure on the sigma-algebra $mathcal{A}$?

I tried it with:

$mu_f(emptyset)=int_emptyset f dmu = 0$.

I'm not sure if this is right.

For the countable additivity I don't know how to show that

$mu_f(cup^{i=1}_{infty}E_i)=sum_{i in I}{mu_f(E_i)}$.

$$mu_f(varnothing)=int_{varnothing}f;dmu=intmathbf1_{varnothing}f;dmu=int0;dmu=0$$

Further be aware that we always have $intsum_{i=1}^{infty}g_i;dmu=sum_{i=1}^{infty}int g_i;dmu$ if the $g_i$ are measurable and nonnegative.

By disjoint and measurable $E_i$ moreover we have $mathbf1_{bigcup_{i=1}^{infty}E_i}=sum_{i=1}^{infty}intmathbf1_{E_i}$ so that:

$$mu_f(bigcup_{i=1}^{infty}E_i)=int_{bigcup_{i=1}^{infty}E_i}f;dmu=intmathbf1_{bigcup_{i=1}^{infty}E_i}f;dmu=intsum_{i=1}^{infty}mathbf1_{E_i}f;dmu=sum_{i=1}^{infty}intmathbf1_{E_i}f;dmu=$$$$sum_{i=1}^{infty}mu_f(E_i)$$