What is the measure of $(\infty,\infty)$

Question

I'm dealing with a counter-example. Specifically, let $E = \cap_{k=1}^{\infty}E_k$ and $E_{k+1} \subset E_k$. We have

$$m(E) = \lim_{n \to \infty} m(E_n)$$

only if $m(E) < \infty$. My book says $E_n = (n, \infty)$ is a counter-example when $m(E)$ is not less than $\infty$. $E = \cap_{n=1}^{\infty}E_n$ is empty so it has measure zero. Then what is $\lim_{n \to \infty} m(E_n)$? The way that I think about it is that you need to cover $\lim_{n \to \infty} E_n$ with $(\infty, \infty)$, and the measure of $(\infty, \infty)$ is $\infty$? Is that correct?

Answer 1

Let $M>0$ be arbitrary. Then $m(E_n) > m((n, n+M)) = M$. Since this is true for any $n$,

$$\lim_{n\to\infty} m(E_n) =\infty$$

Answer 2

$m(E_n)=\infty$ for each $n$ so $\lim m(E_n)=\infty$.

Answer 3

In this context, $\infty$ is a well-defined mathematical object that comprises a single point at the "end" of the real number line - that is, for every other real number $x$, $x < \infty$. The same total ordering properties apply as before, and thus there are no "numbers" strictly between $\infty$ and itself. This means $(\infty, \infty)$ is an empty interval, thus its measure is $0$.

On the other hand, the measure of every interval $(x, \infty)$ for finite $x$ is $\infty$. So the limit of measures as $x$ goes to $\infty$ is $\infty$, however the measure of the "final" interval $E = (\infty, \infty)$ is, as above, $0$, thus a counterexample for the case where the "approximating" intervals do have infinite measure.