Suppose we have a measurable space $(X, A)$, $\{\mu_n\}_{n\geq1}$ is a sequence of increasing measures on it. (i.e. $\mu_n(E) \leq \mu_{n+1}(E), \forall E \in A$).
First prove $\mu(A)= \sup(\mu_n(A))$ is a measure.
Then prove for non-negative measurable function $f$, we have,
$$\int f d\mu=\sup_{n\geq 1}\int f d\mu_n$$.
My try and questions:
Prove $\mu$ is sigma additive. $\mu(\cup E_i)=\sup_{n} \mu_n(\cup E_i)= \lim _n\mu_n(\cup E_i)= \lim_{n} \sum \mu_n(E_i)$, then how can we change the limits and summation?
For simple functions, I can easily check $\int f d\mu=\sup_{n\geq 1}\int f d\mu_n$, however when dealing with non-negative functions, I met problems. Suppose $f_m$ is the sequence of simple functions that converge to $f$, so we have $\int f d\mu= \lim_{m}\int f_m d\mu = \lim_{m} \sup_{n\geq 1}\int f_m d\mu_n$, If I can change supremum and limits, I will get the answer. But I can not make it clear.
(2) Notice that $\mu_n\leq \mu$ for all $n$ and so, $\int f\,d\mu_n\leq \int f\,d\mu$ for all $n$ and all measurable $f\geq0$. Equality holds for all simple functions (by definition of $\mu$).
Suppose $f\geq0$ is measurable. Let $f_m$ a sequence of simple nonnegative functions such that $f_m\nearrow f$.
Notice that if $\sup_n\int f\,d\mu_n<\infty$, then $$\int f_m\,d\mu=\sup_n \int f_m\,d\mu_n\leq\sup_n\int f\,d\mu_n<\infty$$ Hence $$\int f\,d\mu=\lim_m\int f_m\,d\mu\leq \sup_n\int f\,d\mu_n<\infty.$$ This also shows that $\int f\,d\mu=\sup_n\int f\,d\mu_n$.
(1) I leave the details to the OP (details to work out are marked by why?).
Notice that
Let $(A_n:n\in\mathbb{N})$ be a sequence of pairwise disjoint measurable functions whose union is $A$. Then $$\mu(A)\geq\mu\Big(\bigcup^n_{j=1}A_j\Big)=\sum^n_{j=1}\mu(A_j)$$ Hence $\mu(A)\geq\sum^\infty_{n=0}\mu(A_n)$. Now, if $c<\mu(A)$, then there is $N$ such that $c<\mu_N(A)$ (why?). It follows that $$c<\mu_N(A)=\sum_k\mu_N(A_k)\leq\sum_k\mu(A_k)$$ As this holds for all $c<\mu(A)$, we can conclude that $$\mu(A)=\sum_k\mu(A_k)$$