How do I prove that this function is a measure? I have the following problem:

Let $(\Omega, \mathfrak{A}, \mu)$ be a measurespace and let $f:\Omega \rightarrow [0,+\infty]$ be a nonegative simple function, i.e. $f(\Omega)=\{b_1,...b_s\}$. We define for $a\in \mathfrak{A}$ $$\int_A fd\mu=\int_\Omega \chi_A f d\mu$$. Show that the function $$A\mapsto \int_A fd\mu$$ is a measure.

My idea was the following:
Proof
We need to check the following two points. First it is easy to see that $$\emptyset \mapsto \int_\emptyset f d\mu=0$$But now let $I$ be a countable index set and let $\{A_i\}_{i\in I}\in \mathfrak{A}$ such that $A_i \cap A_j=\emptyset$ if $i\neq j$. Now consider $$\bigcup_{i=1}^\infty A_i \in \mathfrak{A}$$ and look at $$\int_{\bigcup_{i=1}^\infty A_i} f d\mu=\int_\Omega f \chi_{\bigcup_{i=1}^\infty A_i} d\mu\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$Let us remark that $$\chi_{\bigcup_{i=1}^\infty A_i}(x)\leq \sum_{i=1}^\infty \chi_{A_i} (x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$ but since the $A_i$'s are disjoint we have indeed $$\chi_{\bigcup_{i=1}^\infty A_i}=1 \Leftrightarrow x\in A_i$$ for some $i\in I$. Therefore from this fact we have equality in $(2)$.Then we have $$\int_\Omega f \chi_{\bigcup_{i=1}^\infty A_i} d\mu=\int_\Omega f \cdot \sum_{i=1}^\infty \chi_{A_i}d\mu=\int_\Omega \left(\sum_{i=1}^\infty f\chi_{A_i}\right) d\mu$$. But now since we can swap the integrals and the sum we get $$\int_\Omega \left(\sum_{i=1}^\infty f\chi_{A_i}\right) d\mu=\sum_{i=1}^\infty \int_\Omega f\chi_{A_i} d\mu=\sum_{i=1}^\infty \int_{A_i} f d\mu\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$Now putting $(1)$ and $(3)$ together gives the sigma additivity.
I'm not sure if this works like this. could someone have a look maybe?
Thanks for your help
 A: The structure of your proof is correct, but it has huge leaps that need explanation. First, you give the inequality
$$\mathbf{1}_{\cup_iA_i}(x)\leq\sum_i \mathbf{1}_{A_i}(x)$$
If you're going to state this fact, you must explain why this holds for any countable collection of sets. You also need to show why, in the case that the $A_i$ s are disjoint, this inequality becomes an equality. Another massive gap is

But now since we can swap the integrals and the sum...

WHY?? Why can we swap the integral and the sum? Big hint - use the fact that $f$ is non-negative and simple!
To help you guide your thinking, here is an alternate proof. To avoid messy subscripts I will write the indicator function of the set $S$ as $\mathbf{1}(S)$. I.e $\mathbf{1}(S)(x)=1$ for $x\in S$ and $0$ otherwise.

Let $(X,\Sigma,\mu)$ be a measure space and let $\varphi$ be a non-negative simple function:
$$\varphi=\sum_{i=1}^n a_i\mathbf{1}(E_i)$$
Define the mapping
$$\nu(A)=\int_{A}\varphi\mathrm d\mu=\int_X \varphi \cdot \mathbf{1}(A)\mathrm d\mu$$
Let's show $\nu$ is a measure on $(X,\Sigma)$. The requirements $\nu:\Sigma\to [0,\infty]$ and $\nu(\emptyset)=0$ are very easily checked, so let's now check countable disjoint additivity. First, we can remark that
$$\nu(A)=\int_X \varphi\cdot \mathbf{1}(A) \mathrm d\mu = \int_X \sum_{i=1}^n\left(\mathbf{1}(A)\cdot a_i\mathbf{1}(E_i)\right) \mathrm d\mu$$
It is very easily checked that for any two sets $A,B$ that $\mathbf{1}(A)\cdot\mathbf{1}(B)=\mathbf{1}(A\cap B)$, and hence
$$\nu(A)=\int_X \sum_{i=1}^na_i\mathbf{1}(A\cap E_i)\mathrm d\mu$$
Since this is a finite sum, we have no problems interchanging summation with integration.
$$\nu(A)=\int_X \sum_{i=1}^na_i\mathbf{1}(A\cap E_i)\mathrm d\mu=\sum_{i=1}^n a_i \int_X \mathbf{1}(A\cap E_i)\mathrm d\mu=\sum_{i=1}^n a_i\mu(A\cap E_i)$$
Since the integrand is simple. So, let $\{F_j\}_j$ be a countable collection of disjoint sets. Then
$$\nu\left(\cup_j F_j\right)=\sum_{i=1}^na_i\mu\big(E_i\cap (\cup_jF_j)\big)$$
Using elementary set algebra,
$$E_i\cap(\cup_j F_j)=\cup_j(E_i\cap F_j)$$
Thus
$$\nu(\cup_j F_j)=\sum_{i=1}^na_i\mu\big(\cup_j(E_i\cap F_j)\big)$$
Since the sets $\{F_j\}_j$ are disjoint, the sets $\{E_i\cap F_j\}_j$ certainly are as well, for any fixed $i$, and since $\mu$ is a measure, it has countable disjoint additivity,
$$\nu(\cup_j F_j)=\sum_{i=1}^na_i\sum_j\mu(E_i\cap F_j)$$
Now we can interchange the order of summation (why?) and hence
$$\nu(\cup_j F_j)=\sum_j\underbrace{\sum_{i=1}^na_i\mu(E_i\cap F_j)}_{=\nu(F_j)} \\ \nu(\cup_j F_j)=\sum_j\nu(F_j)$$
Done.
