Measure of a ray

Question

Consider $E_n = [n, \infty)$. What is than the limit: $$\lim_{n \to \infty}\mu(E_n)$$ Is it $0$ or $\infty$?
On one hand, $\forall n: \mu({E_n}) = \infty$. On the other hand, consider $x \in \mathbb{R}$, for each $x$ there exist N such that $x \notin E_n \; \forall n \geq N$.

Answer 1

I still disagree with commentators who state this limit doesn't exist or "is not well defined". The answer is that $lim_{n \to \infty} \mu (E_n) = \infty$, and it's not a random number, it's just a limit of a stationary sequence. The quality mentioned in comments: $\mu(\lim E_n) = \lim \mu (E_n)$ only holds for finite measure sets, as I said, so it doesn't contraidict anything.

To clear things up: I consider space [$-\infty, +\infty$]. I thought it's obvious, since I asked about $+\infty$ sequences limit, but still.

My previous answer was hidden by people who didn't even bother to answer my question, but whatsoever I consider this post as an answer and will mark it so in two days.

Answer 2

Maybe to precisely phrase the setting you are working in, I think you are considering the topological space $[0,\infty]$ with the order topology. That is $[0,\infty)\sqcup\{ \infty \}$ with a topology generated by rays $(a,\infty)\sqcup\{\infty\}$ and $(b,\infty)$ or $[0,\infty)$.

Within this space $\mu(E_n)\equiv \infty$ and so the limit must be $\infty$, as this space is Hausdorff. This is essentially what julio_es_sui_glance told you.