Let $\mu_1,\mu_2,\dots$ measures on the measurable space $(E,\mathcal{E})$. Then $\mu = \sum_n \mu_n$ is a measure on $(E,\mathcal{E})$.
I want to show, that $$\mu f = \sum_n \mu_n f$$ for every positiv measurable function $f \in \mathcal{E}_+$ where $\mu f = \int_E f \, \mathrm{d}\mu$.
My strategy is to show for the function $L \colon \mathcal{E}_+ \to [0,\infty]$, $L(f)=\sum_n \mu_n f$ that the following properties hold:
- $f=0 \Rightarrow L(f) = 0$
- $f,g \in \mathcal{E}_+$, $a,b \in \mathbb{R}_+$ $\Rightarrow$ $L(af+bg)=aL(f)+bL(f)$
- $(f_k) \subset \mathcal{E}_+$, $f_k \nearrow f$ $\Rightarrow$ $L(f_k) \nearrow L(f)$
Then there exists a unique measure $\nu$ on $(E,\mathcal{E})$ such that $L(f)=\nu f$ for every $f \in \mathcal{E}_+$ and because $ \nu(B) = L(1_B) = \mu(B), \ B \in \mathcal{E} $ the statement $\nu = \mu$ is valid.
My problem with this is the third point:
For every $N \in \mathbb{N}$
$$ \sup_k \sum_{n=1}^{N} \mu_n(f_k) = \sum_{n=1}^{N} \sup_k \mu_n(f_k) = \sum_{n=1}^{N} \mu_n(f) $$ from the monotone convergence theorem.
Now I wan't to interchange $\lim$ and $\sup$ like this:
$$ \sum_n \mu_n (f) = \lim_{N \to \infty}\sup_k \sum_{n=1}^{N} \mu_n(f_k)= \sup_k \lim_{N \to \infty} \sum_{n=1}^{N} \mu_n(f_k) = \sup_k \sum_{n} \mu_n(f_k) $$
Is this possible and why?
Concerning the issue with the sup, don't forget that the supremum in $k$ there is also the limit in $k$. Hence the monotone convergence theorem applies. To see this, define functions $a_{k} : \mathbb{N} \to [0,\infty]$ by \begin{equation*} a_{k}(n) = \int_{E} f_{k} \, d \mu_{n} \end{equation*} Notice that $a_{k}(n) \leq a_{k+ 1}(n) \leq \int_{E} f \, d \mu_{n}$ and $\lim_{k \to \infty} a_{k}(n) = \int_{E} f \, d \mu_{n}$. Therefore, by the monotone convergence theorem (applied to $\mathbb{N}$ with the counting measure), \begin{equation*} \lim_{k \to \infty} \sum_{n = 1}^{\infty} a_{k}(n) = \sum_{n = 1}^{\infty} \int_{E} f \, d \mu_{n}. \end{equation*}
On the other hand, one can check directly that the formula $\mu(A) = \sum_{n = 1}^{\infty} \mu_{n}(A)$ defines a measure on $\mathcal{E}$. Further, if $f$ is a simple function, then $\int_{E} f \, d \mu = \sum_{n = 1}^{\infty} f \, d \mu_{n}$ follows immediately. Arguing using the monotone convergence theorem (as above) and simple approximation, one then finds that the same identity holds when $f$ is bounded and non-negative. From there, one can generalize to arbitrary non-negative or integrable $f$.