Let $\mu$ be a measure on $(X, \mathcal{A})$ and a measurable function $f:X \to \mathbb{R}, \ f \geq 0$.
Define $\mu_f(E): \mathcal{A} \to \mathbb{R}, \ \mu_f(E):=\int_E f \ d\mu$ for $E \in \mathcal{A}$.
How to prove that $\mu_f$ is a measure on the sigma-algebra $\mathcal{A}$?
I tried it with:
$\mu_f(\emptyset)=\int_\emptyset f \ d\mu = 0$.
I'm not sure if this is right.
For the countable additivity I don't know how to show that
$\mu_f(\cup^{i=1}_{\infty}E_i)=\sum_{i \in I}{\mu_f(E_i)}$.
Best Answer
$$\mu_f(\varnothing)=\int_{\varnothing}f\;d\mu=\int\mathbf1_{\varnothing}f\;d\mu=\int0\;d\mu=0$$
Further be aware that we always have $\int\sum_{i=1}^{\infty}g_i\;d\mu=\sum_{i=1}^{\infty}\int g_i\;d\mu$ if the $g_i$ are measurable and nonnegative.
By disjoint and measurable $E_i$ moreover we have $\mathbf1_{\bigcup_{i=1}^{\infty}E_i}=\sum_{i=1}^{\infty}\mathbf1_{E_i}$ so that:
$$\mu_f(\bigcup_{i=1}^{\infty}E_i)=\int_{\bigcup_{i=1}^{\infty}E_i}f\;d\mu=\int\mathbf1_{\bigcup_{i=1}^{\infty}E_i}f\;d\mu=\int\sum_{i=1}^{\infty}\mathbf1_{E_i}f\;d\mu=\sum_{i=1}^{\infty}\int\mathbf1_{E_i}f\;d\mu=$$$$\sum_{i=1}^{\infty}\mu_f(E_i)$$