Let $\left(\mu_n\right)_{n=1}^\infty$ be a sequence of Radon measures on $X\subset \mathbb{R}^n$ such that for every compact subset $K\subset X:\sup_{n=1,2,\dots}\mu_n(K) < \infty$. Is it then true that $\tilde{\mu}(A)\equiv \lim\sup_{n\to\infty}\mu_n(A)$ is at least a Radon outer measure?
(Thoughts:) I got stuck trying to prove that the limit supremum is countable subadditive and that each Borel set $B\subset X$ is in fact measurable. My main question is regarding the 2.) below, as that seems to pose a serious threat to the claim, but I'd still like to hear about your thoughts on 1.) as well. My current feeling is that the claim is not true, for the Borel sets might not be measurable in the limit supremum. But without a further ado, let $(S_i)_{i=1}^\infty$ is a sequence of subsets of $X$. Then,
1.) (Countable subadditivity)
$$\lim\sup_{n\to\infty}\mu_n\left(\bigcup_{i=1}^\infty S_i\right) \leq \lim\sup_{n\to\infty}\sum_{i=1}^\infty \mu_n(S_i) \leq \sum_{i=1}^\infty \lim\sup_{n\to\infty}\mu_n(S_i)$$
where $(S_i)_{i=1}^\infty$ is a sequence of subsets of $X$ and the the first equality is by the virtue that $\mu_n$ is a Radon measure.
2.) (Measurability of Borel sets) Now, let $B\subset X$ be a Borel set. Then, we'd like to have that $$\lim\sup_{n\to\infty}\left(\mu_n(E\cap B) + \mu_n(E\cap B^c)\right) = \lim\sup_{n\to\infty}\left(\mu_n(E\cap B)\right) + \lim\sup_{n\to\infty}\left(\mu_n(E\cap B^c)\right)$$
As for 1.), I think that the inequality holds by the virtue that $\lim\sup_{n\to\infty}a_n = \inf_{m\to\infty}\sup_{n\geq m}a_n$ for a real sequence $(a_n)$ and for any $m\in \mathbb{N}$ we have that $\sup_{n\geq m}\sum_{i=1}^\infty \mu_n(A_i) \leq \sum_{i=1}^\infty \sup_{n\geq m}\mu_n(A_i)$, but I can't say for sure as I'm not that comfortable with infimum juggling .
As for 2.), if the equality does hold, it requires the use of some measure specific tools since it is my understanding that $\lim\sup_{n\to\infty}\left(a_n + b_n\right) \leq \lim\sup_{n\to\infty}\left(a_n\right) + \lim\sup_{n\to\infty}\left(b_n\right)$ for any real sequences $(a_n), (b_n)$.
(Bonus: Weak convergence) I am aware that under the assumptions on the sequence $(\mu_n)$, there exists a Radon measure $\mu$ and a subsequence of $(\mu_{n_k})$ such that $(\mu_{n_k})$ converges weakly to $\mu$, i.e.
$$\forall f \in C_0(X):\lim_{k\to\infty}\int_Xfd\mu_{n_k}=\int_Xfd\mu$$
As of now I am not sure how this fact could be exploited, or whether any additional weak convergence assumption could be used on the original sequence $(\mu_n)$. I decided to include this fact to my post as it seems related, although I can't really comment on its implications.
Take $X = \mathbb{R}$ and define $\mu_k = \delta_0$ if $k$ is even and $\mu_k = \delta_1$ otherwise (where $\delta_0$ and $\delta_1$ denote the Dirac measures at $0$ and $1$). Then if we set $\tilde{\mu}(A) = \limsup_{k \to \infty} \mu_k(A)$ for any Borel set $A \subset \mathbb{R}$ we have $$ \tilde{\mu}(\{0\}) = \tilde{\mu}(\{1\}) = \tilde{\mu}(\{0, 1\}) = 1$$ which implies that $\tilde{\mu}$ is not a Borel measure.