So, I have seen questions that are similar to this one but often Fatou's lemma or the Caratheadory theorem is employed. I would like to prove this only through the definition of the outer measure, which is given as follows;
An outer-measure $m^*$ on a set $X$ assigns to every subset of X (including $+\infty$) a nonnegative number, such that
1) $m^*(\emptyset) =0$
2) $m^*(S) \leq m^*(T) $ when $S \subseteq T \subseteq X$
3) $m^*(\cup^{\infty}_{i=1}S_i) \leq \Sigma_{i=1}^{\infty}m^*(S_i)$ whenever $S_i \subseteq X$.
My question is:
Let $\mu_1^*, \mu_2^*,...$ be a sequence of outer measures on $\mathbb{R}$, such that $\mu^*_n(E) \leq u^*_{n+1}(E)$ for every $E \subseteq \mathbb{R}$, define $\mu^*(E) = lim_{n\rightarrow \infty} \mu^*_n(E) $. Prove that $\mu^*$ is an outer measure.
I have proven the first two parts of the measure definition and would greatly appreciate someone to help me do the last part!
Let $S_i\subseteq \mathbb{R}$, $i\in\mathbb{N}$. Then we have for each $n\in\mathbb{N}$: $$\mu_n(\bigcup_{i\in\mathbb{N}}S_i)\leq\sum_{i\in\mathbb{N}}\mu_n(S_i)\leq\sum_{i\in\mathbb{N}}\mu(S_i).$$ We can therefore conclude $$\mu(\bigcup_{i\in\mathbb{N}}S_i)=\lim_{n\to\infty}\mu_n(\bigcup_{i\in\mathbb{N}}S_i)\leq\sum_{i\in\mathbb{N}}\mu(S_i).$$