# Measure on a sigma-algebra with integral

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

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

$$\mu_f(\varnothing)=\int_{\varnothing}f\;d\mu=\int\mathbf1_{\varnothing}f\;d\mu=\int0\;d\mu=0$$
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:
$$\mu_f(\bigcup_{i=1}^{\infty}E_i)=\int_{\bigcup_{i=1}^{\infty}E_i}f\;d\mu=\int\mathbf1_{\bigcup_{i=1}^{\infty}E_i}f\;d\mu=\int\sum_{i=1}^{\infty}\mathbf1_{E_i}f\;d\mu=\sum_{i=1}^{\infty}\int\mathbf1_{E_i}f\;d\mu=$$$$\sum_{i=1}^{\infty}\mu_f(E_i)$$