Outer Lebesgue Measure limit's proof

[SqueeSpleen]

If $\{ E_{k} \}_{k \in \mathbb{N}}$ is an increasing sequence of subsets of $R^{p}$, then:
$| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} = \lim_{k \to \infty} |E_{k}|_{e}$
I proved:
$| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} \geq \lim_{k \to \infty} |E_{k}|_{e}$
But I don't know how to prove the other inequality.

$E_{n} = \displaystyle \bigcup_{k=1}^{n} E_{k} \subseteq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}$
So, for monotony of the outer Lebesgue measure we have:
\newline
$| E_{n} |_{e} = | \displaystyle \bigcup_{k=1}^{n} E_{k} |_{e} \leq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}$
\newline
So the limit has also to be equal or lesser.
My problem is that I have not idea how to link the limit of the sets with the other limit in the other way, I could decompose:
$\displaystyle \bigcup_{k=1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle \bigcup_{k=n+1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle (\bigcup_{k=n+1}^{\infty} E_{k} - \bigcup_{k=1}^{n} E_{k})$ but the outer measure is subaditive, so the decomposition of the measure I could do to try to get something arbitrary small would be greater than my previous set and I have no guarantee it's close than the measure of the other set (vitali's could basically double the external measure I'm trying to delimite).
If they're measurables it's very easy, but in general I don't know how to prove it.

 

[gopher_p]

If $\{ E_{k} \}_{k \in \mathbb{N}}$ is an increasing sequence of subsets of $R^{p}$, then:



$| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} = \lim_{k \to \infty} |E_{k}|_{e}$
I proved:
$| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} \geq \lim_{k \to \infty} |E_{k}|_{e}$
But I don't know how to prove the other inequality.

$E_{n} = \displaystyle \bigcup_{k=1}^{n} E_{k} \subseteq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}$
So, for monotony of the outer Lebesgue measure we have:
\newline
$| E_{n} |_{e} = | \displaystyle \bigcup_{k=1}^{n} E_{k} |_{e} \leq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}$
\newline
So the limit has also to be equal or lesser.
My problem is that I have not idea how to link the limit of the sets with the other limit in the other way, I could decompose:
$\displaystyle \bigcup_{k=1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle \bigcup_{k=n+1}^{\infty} E_{k} = \displaystyle \bigcup_{k=1}^{n} E_{k} \cup \displaystyle (\bigcup_{k=n+1}^{\infty} E_{k} - \bigcup_{k=1}^{n} E_{k})$ but the outer measure is subaditive, so the decomposition of the measure I could do to try to get something arbitrary small would be greater than my previous set and I have no guarantee it's close than the measure of the other set (vitali's could basically double the external measure I'm trying to delimite).
If they're measurables it's very easy, but in general I don't know how to prove it.

Well the first thing I would do is split the problem into two cases; one where $|\bigcup_{k=1}^{\infty} E_{k} |_{e}$ is finite, and one where it is not.

For the harder (finite) case, I think you might just need to roll up your sleeves, dust off the old $\epsilon$s and $\delta$s (or I guess in this case $\epsilon$s and $n$s), and play around in the muck for a bit. Maybe draw a picture/cartoon of a "nice-looking" finite case, come up with an heuristic argument you think works there, and then see if you can give that argument some rigor that makes it work in the general case.