Increasing limit of measures is a measure

Question

I wonder if someone can check if this proof is correct?

Let $(X,\mathcal{F})$ be a measure space and $\{\mu_n\}_n\in \mathbb{N}, \mu_n: \mathcal{F} \to \mathbb{R}^+$ be an increasing sequence of measures $(\forall A \in \mathcal{F}, \forall n, \mu_n(A)\leq \mu_{n+1}(A))$. Show that $\mu:=\lim_{n\to \infty}\mu_n $ is a measure on $(X,\mathcal{F})$.

1) $\mu(\emptyset)= \lim_{n\to \infty}\mu_n(\emptyset)=\lim_{n\to \infty}0=0$

2) Let $A, B \in \mathcal{F}$ be disjointsets. Then $\mu(A \cup B)= \lim_{n\to \infty}\mu_n(A \cup B)= \lim_{n\to \infty}(\mu_n(A) + \mu_n(B))= \lim_{n\to \infty}\mu_n(A) + \lim_{n\to \infty} \mu_n(B)= \mu(A) +\mu(B)$. So $\mu$ is finitely additive.

3) I want to show that $\mu$ is $\sigma-$additive. Let $( A_i )_{i\geq1}$ be a sequence of disjoint sets in $\mathcal{F}$. $|\sum_{i=1}^N \mu(A_i) - \mu(\cup_{i=1}^{\infty}A_i)|\leq |\sum_{i=1}^N \mu(A_i) -\sum_{i=1}^N \mu_n(A_i)+ \mu_n(\cup_{i=1}^{\infty}A_i) -\mu(\cup_{i=1}^{\infty}A_i)| \leq \sum_{i=1}^N |\mu(A_i) -\mu_n(A_i)|+ |\mu_n(\cup_{i=1}^{\infty}A_i) -\mu(\cup_{i=1}^{\infty}A_i)|$

Choose $n$ so large that $\forall i \ \ |\mu(A_i) -\mu_n(A_i)|< \epsilon2^{-i}$ and $|\mu_n(\cup_{i=1}^{\infty}A_i) -\mu(\cup_{i=1}^{\infty}A_i)|<\epsilon$. Then $|\sum_{i=1}^N \mu(A_i) - \mu(\cup_{i=1}^{\infty}A_i)|<2\epsilon$, so $\lim_{N\to\infty}\sum_{i=1}^N \mu(A_i) = \mu(\cup_{i=1}^{\infty}A_i)$.

I wonder if the last part(3) is correct? I know that this can be shown in a different way using the fact that "$\sigma$-additive" $\iff$ "continuous from below", or using the monotone convergence theorem but I'd prefer avoiding that.

Answer 1

Your 3) runs into subtraction trouble on sets of infinite $\mu$ measure.

To avoid this, use the fact that it's okay to interchange monotone inceasing limits. Let $A=\cup_nA_n$ be a disjoint union of $\mathcal F$ sets. Then $$ \eqalign{ \mu(A) &=\lim_k\mu_k(A)=\lim_k\sum_{n=1}^\infty\mu_k(A_n)\cr &=\lim_k\lim_N\sum_{n=1}^N\mu_k(A_n)\cr &=\lim_N\lim_k\sum_{n=1}^N\mu_k(A_n)\cr &=\lim_N\sum_{n=1}^N\lim_k\mu_k(A_n)\cr &=\lim_N\sum_{n=1}^N \mu(A_n) =\sum_{n=1}^\infty\mu(A_n).\cr } $$

Answer 2

(Seems to work just for probability measures.) Let $\mu_1' = \mu_1$ and $\mu_n' := \mu_{n+1}-\mu_n$ for $n>1$. (with the rule $\infty - \infty := \infty$). Note that $\mu_n'$ is a measure for all $n\in\mathbb{N}$, and that $\mu_{\infty}(A) = \sum_{i=1}^{\infty} \mu_i'(A)$ for all $A\in\mathcal{F}$. Then using your disjoing $A_i$ with $A:=\bigcup_{i=1}^{\infty} A_i$ we can compute

$$\sum_{i=1}^{\infty}\mu_{\infty}(A_i) = \sum_{i=1}^{\infty}\sum_{j=1}^{\infty} \mu_j'(A_i) =\sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \mu_j'(A_i) = \sum_{j=1}^{\infty} \mu_{j}'(A) = \mu_{\infty}(A), $$ since we can always rearrange sums of non-negative summands.