$\lim_{n \to \infty}[n,+\infty)$ has Lebesgue measure 0? I'm trying to show that $\lim_{n \to \infty}[n,+\infty)$ has Lebesgue measure 0. I'm so tempted to just state that as $n \to \infty$, the interval gets smaller and smaller and eventually to 0 but I know for sure that is not a legal move. Any hint is appreciated!
 A: First off, the notation $\lim_{n\to\infty} A_n$ where $A_n$ are subsets of a larger space is not really well-defined.  What is the limiting procedure?  How do you compute such a limit?  In high-faluten language, in what category are you taking the limit?
That being said, a the usual interpretation is as follows: we first define the "limit superior" and "limit inferior" of a sequence of sets.  These are
$$
\limsup A_n := \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k
\qquad\text{and}\qquad
\liminf A_n := \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k.
$$
Basically, the limit superior is the collection of point that appear in infinitely many of the sets, and the limit inferior is the set of points that appear in all but finitely many of the sets.  If
$$ \limsup A_n = \liminf A_n = A, $$
then it is reasonable to define $\lim_{n\to\infty} A_n = A$.  In the case where $A_n = [n,\infty)$, the $A_n$ are a monotonically decreasing family of sets (that is, $A_{n+1} \subseteq A_n$ for all $n\in\mathbb{N}$), and so these definitions reduce to an intersection:
$$ \limsup [n,\infty)
= \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} [k,\infty)
= \bigcap_{n=1}^{\infty} [n,\infty)
= \emptyset $$
and
$$ \liminf [n,\infty)
= \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} [k,\infty)
= \bigcup_{n=1}^{\infty} \emptyset
= \emptyset. $$
To see that the intersection really is empty, it is sufficient to show that if $x \in \mathbb{R}$, then $x \not\in \bigcap [n,\infty)$.  For this, it is sufficient to show that $x \not\in [n,\infty)$ for at least one $n$.  But this is straightforward:  if $n$ is an integer with $n > x$, then $x \not\in [n,\infty)$.  Since we can always find an integer bigger than $x$, it must be that $x\not\in \bigcap [n,\infty)$.  Therefore if $x \in \mathbb{R}$, then
$$ x\not\in \bigcap_{n=1}^{\infty} [n,\infty). $$
Then, as claimed, $\bigcap [n,\infty) = \emptyset$, from which it follows that it has zero Lebesgue measure.
