- #1

SqueeSpleen

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

[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} = \lim_{k \to \infty} |E_{k}|_{e}[/itex]

I proved:

[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} \geq \lim_{k \to \infty} |E_{k}|_{e}[/itex]

But I don't know how to prove the other inequality.

[itex]E_{n} = \displaystyle \bigcup_{k=1}^{n} E_{k} \subseteq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]

So, for monotony of the outer Lebesgue measure we have:

\newline

[itex]| E_{n} |_{e} = | \displaystyle \bigcup_{k=1}^{n} E_{k} |_{e} \leq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]

\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:

[itex]\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})[/itex] 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.

[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} = \lim_{k \to \infty} |E_{k}|_{e}[/itex]

I proved:

[itex]| \displaystyle \bigcup_{k=1}^{\infty} E_{k} |_{e} \geq \lim_{k \to \infty} |E_{k}|_{e}[/itex]

But I don't know how to prove the other inequality.

[itex]E_{n} = \displaystyle \bigcup_{k=1}^{n} E_{k} \subseteq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]

So, for monotony of the outer Lebesgue measure we have:

\newline

[itex]| E_{n} |_{e} = | \displaystyle \bigcup_{k=1}^{n} E_{k} |_{e} \leq \displaystyle \bigcup_{k=1}^{\infty} E_{k} \forall n \in \mathbb{N}[/itex]

\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:

[itex]\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})[/itex] 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.

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