Convergence of a sequence of sets

  • #1
benorin
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Summary:

I need a little help with Baby Rudin material regarding the convergence of a sequence of sets please

Main Question or Discussion Point

I need a little help with Baby Rudin material regarding the convergence of a sequence of sets please. I wish to follow up on this thread with a definition of convergence of a sequence of sets from Baby Rudin (Principles of Mathematical Analysis, 3rd ed., Rudin) pgs. 304-305:

(pg. 304) Definition 11.7) Let ##\mu## be additive, regular, nonnegative, and finite on the ring of elementary sets. Consider countable coverings of any set ##E\subset \mathbb{R}^p## by open elementary subsets ##A_n## :

$$E\subset \bigcup_{n=1}^{\infty}A_n$$

Define

(17) $$\mu^* (E) = \inf \sum_{n=1}^\infty \mu (A_n)$$

the inf being taken over all countable coverings of ##E## by open elementary subsets. ##\mu ^* (E)## is called the outer measure of E, corresponding to ##\mu##.

(pg. 305) Definition 11.9) For any ##A\subset \mathbb{R}^p, B\subset \mathbb{R}^p##, we define

(22)$$S(A,B)=(A-B)\cup (B-A)$$
(23)$$d(A,B)=\mu^* (S(A,B))$$

We write ##A_n\rightarrow A## if

$$\lim_{n\rightarrow\infty} d(A_n , A) = 0$$

With this notion of limit of a sequence of sets, I wish to do what was conveyed in the post linked in the second sentence from the top of this post, namely I wish to use the Lebesgue's Dominated Convergence Theorem to establish that for some given ##A_n , A\subset \mathbb{R}^p## such that ##A_n\rightarrow A##

(eqn 1) $$\lim_{n\rightarrow\infty}\iint_{A_n}\cdots \int g(\vec{x})d\vec{x} = \iint_{A}\cdots \int g(\vec{x})d\vec{x}$$

Of particular interest is the case of ##S_n := \left\{ (x_1 , x_2 , \ldots , x_N ) : \sum_{k=1}^{N}x_k^{2n}\leq 1\right\}##, I need help on how to show that ##S_n\rightarrow S## for ##S:=\left\{ (x_1 , x_2 , \ldots , x_N ) | -1\leq x_k \leq 1, k=1,2,\ldots , N\right\}##? It's been 20+ years since I've done analysis so any help you can give on how to prove (eqn 1) or help on how to show that ##S_n\rightarrow S## is appreciated. Thanks!
 
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Answers and Replies

  • #2
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I have no solution, but a strategy. If we first list what we have, i.e. the definitions, then we get
\begin{align*}
S_n \longrightarrow S &\Longleftrightarrow d(S_n,S)\longrightarrow 0\\
d(S_n,S)&= \mu^*(\;S(S_n-S)\;)\\&=\mu^*(\;(S_n-S)\cup(S-S_n)\;)\\&=\inf\left\{\,\sum \mu(\;(S_n-S)\cup(S-S_n)\;)\,\right\}
\end{align*}
and with ##S_n\subseteq S## we have to show that ##\inf \left\{\sum \mu(\;S-S_n\;)\right\}=0##.

Next we draw an image. This restricts us on two dimensions and small ##n##, but it gives an idea.
For ##N=2## and ##n=4## we get:

1579618840554.png


where ##S-S_n## is the read area. Growing ##n## makes the blue "circle" more "quadratic". So it is sufficient to show that the maximal distance, between the yellow and the black dot, converges to zero. Or better, find a set description of such a red corner and show that its measure converges to zero.
 
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  • #3
benorin
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Since the boundary equation of ##S_n##, namely ##\sum_{k=1}^{N}x_k^{2n}=1## is symmetric in it's variables and the same is true for the boundary equation for ##S##, namely ##|x_k |=1## for ##k=1,2,\ldots , N## the maximal distance will occur (assuming ##x_k\geq 0## for ##k=1,2,\ldots , N## because of symmetry) when all variables are equal, so let ##x=x_k## for ##k=1,2,\ldots , N##, then

$$\sum_{k=1}^{N}x_k^{2n}=1\Rightarrow \sum_{k=1}^{N}x^{2n}=Nx^{2n}=1\Rightarrow |x|=N^{-\frac{1}{2n}}$$

similarly on the boundary of ##S## we have ##(1,1,\ldots ,1)## as the indicated point so the square of the maximal distance then becomes

$$d^2(x)=\sum_{k=1}^N\left( 1- x\right) ^2 = \sum_{k=1}^N\left( 1- N^{-\frac{1}{2n}}\right) ^2 = N\left( 1- N^{-\frac{1}{2n}}\right) ^2$$

Hence ##d^2(x)\rightarrow 0+## as ##n\rightarrow\infty## and thus ##S_n\rightarrow S##.
 
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  • #4
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Yes, that's correct. Now we need such an argument for ##\mu## and some covering of the red areas, because we have some abstract outer measure and not the Euclidean one. It seems as if we could use a finite subcover of
$$
E_n=\bigcup_{x\in \partial S} U(x;r_n)\text{ with } r_n=\dfrac{1}{n}+N(1-N^\frac{1}{2n})^2
$$
which exists by compactness of ##S##. The ##U(x;r_n)=\{\,y\in \mathbb{R}^N\,:\,\|x-y\|<r_n\,\}## are the open balls around the points on the boundary ##\partial S## of ##S##. The additional summand ##\frac{1}{n}## is just a correction, such that we won't have to bother the boundaries, because our sets are closed and the covering should be open.

So formally it remains to show ##d(S_n,S) \stackrel{\mu^*}{\longrightarrow} 0##. I.e. that ##\sum_{k=1}^M \mu(A_k) \stackrel{n\to \infty}{\longrightarrow} 0## if ##\{\,A_1,\ldots,A_M\,\}## are a finite subcover of ##E_n\,.## The ##A_k## and ##M## depend on ##n## so there should be an additional index ##n##, but we only have to show it for one specific ##n##, so I left it out.
 
  • #5
benorin
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Yes, that's correct. Now we need such an argument for ##\mu## and some covering of the red areas, because we have some abstract outer measure and not the Euclidean one. It seems as if we could use a finite subcover of
$$
E_n=\bigcup_{x\in \partial S} U(x;r_n)\text{ with } r_n=\dfrac{1}{n}+N(1-N^{-\frac{1}{2n}})^2
$$
which exists by compactness of ##S##. The ##U(x;r_n)=\{\,y\in \mathbb{R}^N\,:\,\|x-y\|<r_n\,\}## are the open balls around the points on the boundary ##\partial S## of ##S##. The additional summand ##\frac{1}{n}## is just a correction, such that we won't have to bother the boundaries, because our sets are closed and the covering should be open.

So formally it remains to show ##d(S_n,S) \stackrel{\mu^*}{\longrightarrow} 0##. I.e. that ##\sum_{k=1}^M \mu(A_k) \stackrel{n\to \infty}{\longrightarrow} 0## if ##\{\,A_1,\ldots,A_M\,\}## are a finite subcover of ##E_n\,.## The ##A_k## and ##M## depend on ##n## so there should be an additional index ##n##, but we only have to show it for one specific ##n##, so I left it out.
I think you missed the negative in the exponent of N in def. ##E_n## (I added it in the quote).

Let ##\{\,A_1,\ldots,A_M\,\}## be a finite subcover of ##E_n##, hence there exists ##x_1,x_2,\ldots,x_M\in\partial S## such that

$$(S-S_n)\subset\bigcup_{k=1}^{M}A_k=\bigcup_{k=1}^{M}U(x_k;r_n)$$

(where I have totally ignored the fact that ##x_k## may vary with ##n## and ##N##). Here's where I get lost, and thinking back on it too, how to evaluate the norm here, and what I was kicking around thinking back was I used the Euclidian metric back in the last post...
 
  • #6
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We know that the Euclidean volume goes to zero. Now we need an argument, why the ##\mu## volume does, too. The ##E_n## build a filtration, so the properties of ##\mu## should show, that their volume gets smaller and smaller.
 
  • #7
benorin
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Let ##\{\,A_1,\ldots,A_M\,\}## be a finite subcover of ##E_n##, hence there exists ##\vec{x_1},\vec{x_2},\ldots,\vec{x_M}\in\partial S## such that

$$(S-S_n)\subset\bigcup_{k=1}^{M}A_k=\bigcup_{k=1}^{M}U(\vec{x_k};r_n)$$

(where I have totally ignored the fact that ##x_k## may vary with ##n## and ##N##). Define ##G(\vec{x_k};r_n):= \left\{ (x_1,x_2,\ldots,x_N) : |\vec{x}-\vec{x_k}|<r_n\right\}##

$$d(S,S_n)=\mu^* (S-S_n)\leq\mu^* \left( \bigcup_{k=1}^{M}A_k\right) $$
$$= \inf \sum_{k=1}^\infty \mu\left( A_k\right) \leq \sum_{k=1}^{M}\mu ( G(\vec{x_k};r_n))$$
$$ \leq M\max \mu ( G(\vec{x_k};r_n)) $$
$$\rightarrow M\max \mu ( G(\vec{x_k};0+))=0$$

where the inf was taken over all countable coverings of ##S-S_n## by open elementary sets ##A_k## and the limit was as ##n\rightarrow\infty## and the last equality follows from the measure of a countable point set is 0.
 
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  • #8
benorin
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I made a mistake in the OP, namely ##S_n## should be defined as

(correct definition) $$S_n:=\left\{ (x_1,x_2,\ldots,x_N) : \sum_{k=1}^{N}x_k^{2n}\leq N\right\}$$

for compactness' sake because what I originally had was

(original incorrect definition) $$S_n:=\left\{ (x_1,x_2,\ldots,x_N) : \sum_{k=1}^{N}x_k^{2n}\leq 1\right\}$$

and just to make the language simple and clear take the case of N=2 and think on ##\partial S_n## it's missing some of it's limit points so ##S_n## is not even closed hence not compact and we need compactness so I've begun altering our work but I've hit a snag, when I solve for the x coordinates at which the maximal distance occurs I get

$$|x_k| = 1^{\frac{1}{2n} = 1 \, \rm{ for } \, k=1,2,\ldots, N$$

which I think is a problem... thoughts?
 
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  • #9
Infrared
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I don't think the statement you want to prove is true. Let ##A=[0,1]## and ##A_n=[0,1]\cup [n,n+1/n]##. Then ##S(A,A_n)=[n,n+1/n]## has measure ##1/n##, so ##A_n\to A##. Letting ##f(x)=x##, we see ##\int_A f=1/2## but ##\int_{A_n}f=3/2+\frac{1}{2n^2}\longrightarrow 3/2##. (All my integrals are with respect to the Lebesgue measure.)
 
  • #10
benorin
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I'm not sure ##[n,n+\frac{1}{n}]## qualifies as a valid subset of ##\mathbb{R}## as ##n\rightarrow\infty## tho, the text makes no mention of the point at infinity or the extended real number system in the relevant section that I see right now. But you may be correct anyhow. I gambled on this theorem being true so I'm not just gonna give up, but this is concerning to me.
 
  • #11
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I'm not sure what you mean.. ##[n,n+1/n]## is a subset of ##\mathbb{R}## for each ##n##. I'm never literally taking ##n=\infty##.

If you want another example, let ##A=[1/2,1]## and ##A_n=(1/(n+1),1/n)## and ##f(x)=1/x^2.##
 
  • #12
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What we additionally need is a finite, integrable upper bound, ...
I wish to use the Lebesgue's Dominated Convergence Theorem
... to prevent an accumulation at infinity. But this doesn't touch the question about the measures: How do we get from ##\|\cdot\|_2## to ##\mu##.
 
  • #13
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Let ##S,S_n## be as in the problem statement. It is clear that ##S## is the (increasing) union of the ##S_n##. This is already enough to guarantee that ##m(S(S,S_n))=m(S-S_n)## goes to zero as ##n\to\infty##. This is kind of a standard trick: Let ##T_n=S_n-S_{n-1}## (setting ##S_0=\emptyset##). Then ##S=\bigcup_{n=1}^\infty T_n## is a disjoint union, so we can take measures: ##m(S)=\sum_{k=1}^\infty m(T_n)##. Given ##\varepsilon>0## arbitrary, choose ##K## sufficiently large that ##m(S)-\sum_{n=1}^K m(T_n)<\varepsilon##. But ##\sum_{n=1}^K m(T_n)=m(S_K)##, so ##m(S)-m(S_K)<\varepsilon## as desired.

Anyway, in the case of an increasing union like this, the proposition you want does follow from dominated convergence. Let ##\chi,\chi_n## be the indicator functions for ##S,S_n##. Since ##|g(x)|\chi_n\leq |g(x)|\chi##, assuming that ##g## is integrable on ##S##, we can apply the dominated convergence theorem to get

##\lim_{n\to\infty}\int_{S_n}g dm=\lim_{n\to\infty}\int_{\mathbb{R}^p} g\chi_n dm=\int_{\mathbb{R}^p} \lim_{n\to\infty} g\chi_n dm=\int_{\mathbb{R}^p} g\chi dm=\int_{S}g dm.##
 
  • #14
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Let ##S,S_n## be as in the problem statement. It is clear that ##S## is the (increasing) union of the ##S_n##. This is already enough to guarantee that ##m(S(S,S_n))=m(S-S_n)## goes to zero as ##n\to\infty##. This is kind of a standard trick: Let ##T_n=S_n-S_{n-1}## (setting ##S_0=\emptyset##). Then ##S=\bigcup_{n=1}^\infty T_n## is a disjoint union, so we can take measures: ##m(S)=\sum_{k=1}^\infty m(T_n)##. Given ##\varepsilon>0## arbitrary, choose ##K## sufficiently large that ##m(S)-\sum_{n=1}^K m(T_n)<\varepsilon##. But ##\sum_{n=1}^K m(T_n)=m(S_K)##, so ##m(S)-m(S_K)<\varepsilon## as desired.
If we start with such conditions worded for ##m##, sure, there is no problem. My question is, that we have sets defined in an Euclidean norm, and know that e.g. ##\operatorname{Vol}_{eucl.}(A) < \operatorname{Vol}_{eucl.}(B)## since ##A\subsetneq B##, then how could we conclude ##m(A) < m(B) ## and not ##m(A)\leq m(B)?##
 
  • #15
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If we start with such conditions worded for ##m##, sure, there is no problem. My question is, that we have sets defined in an Euclidean norm, and know that e.g. ##\operatorname{Vol}_{eucl.}(A) < \operatorname{Vol}_{eucl.}(B)## since ##A\subsetneq B##, then how could we conclude ##m(A) < m(B) ## and not ##m(A)\leq m(B)?##
I'm not following. Which strictly inequality do I have that you're not sure about?

I just now noticed the change in definition for ##S_n##. I'll try to post an argument for it later.
 
  • #16
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My question is, that given two Euclidean sets as in the problem setup, how can we know that our arbitrary measure inherits the properties of the Euclidean space? For instance: How do we rule out that a Euclidean filtration doesn't become stationary in its ##\mu-##measure.

Or the other way around: How do we know that ##\mu(A) < \mu(B)## only because its Euclidean volumes are? I just don't see the bridge between the Euclidean definition of the sets and their abstract measure ##\mu##. And we may assume ##\mu \neq \lambda##.
 
  • #17
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I'm not using anything specific to Euclidean space. My argument shows that if ##(X,\mathcal{A},\mu)## is a measure space, and ##S=\bigcup_{n=1}^\infty S_n## is an increasing union of measurable sets such that ##\mu(S)<\infty##, then ##\mu(S)-\mu(S_n)=\mu(S-S_n)## has limit ##0##.
 
  • #18
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Yes, I know. But the sets are defined Euclidean. How can we say anything about their connection with ##\mathcal{A}##? The question requires the link, not your argument.

Edit: O.k. if we start with measurability, then o.k. I thought we had to show they are.
 
  • #19
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I don't see what I'm missing. The question is asking to show that ##\lim_{n\to\infty} d(S,S_n)=m(S-S_n)=0##. This is exactly what my argument shows by taking ##\mu## to be the Lebesgue measure ##m##. You have to adjust it a little bit now because the OP changed the definition of ##S_n##, so ##S## is no longer exactly an increasing union of the ##S_n## but other than that I don't see the issue.
 
  • #20
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I don't see what I'm missing. The question is asking to show that ##\lim_{n\to\infty} d(S,S_n)=m(S-S_n)=0##. This is exactly what my argument shows by taking ##\mu## to be the Lebesgue measure. You have to adjust it a little bit now because the OP changed the definition of ##S_n##, so ##S## is no longer exactly an increasing union of the ##S_n## but other than that I don't see the issue.
See my edit. Yes, the change of definition make the sets more inconvenient but doesn't change anything. I was trapped in those Euclidean circles and wondered, how they get a finite measure.
 
  • #21
Infrared
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They're measurable beause they're closed. They have finite measure because they're contained in a box, say ##[-N,N]^p##, of finite measure.
 
  • #22
benorin
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Wow, you guys are writing my paper for me lol! So the theorem holds for the Lebesgue measure (please confirm)? That's good enough for me, thank you. I will post my paper when I'm done typing it up, assuming I can post pdf's here.
 
  • #23
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No, I gave counterexamples in posts 9 and 11
 
  • #24
benorin
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If you want another example, let ##A=[1/2,1]## and ##A_n=(1/(n+1),1/n)## and ##f(x)=1/x^2.##
This is the counter example you gave in #11, your ##A_n## does not converge to ##A##, clearly ##A_n\rightarrow (0+,0+)## which I'm not really sure, as I said it's been 20 years since I've done analysis, would seem to be an open set, yet a singlet, so null set? I'm not sure if there's anything wrong with the other counter example tho. Still thinking (I'm slow)...
 
  • #25
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Sorry, I meant for that to be ##A_n=[1/2,1]\cup (1/(n+1),1/n)##
 

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