How do I prove that this function is a measure?

Solution 1:

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.