Convergence of a sequence of sets

In summary, the Lebesgue's Dominated Convergence Theorem says that if given a sequence of sets, each of which has an outer measure, then the limit of the sequence as n approaches infinity is the set with the largest outer measure.
  • #1
benorin
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I need a little help with Baby Rudin material regarding the convergence of a sequence of sets please
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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  • #2
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
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
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
fresh_42 said:
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
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
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
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
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
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 going to give up, but this is concerning to me.
 
  • #11
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
What we additionally need is a finite, integrable upper bound, ...
benorin said:
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
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
Infrared said:
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
fresh_42 said:
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
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
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
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
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
Infrared said:
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
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
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
No, I gave counterexamples in posts 9 and 11
 
  • #24
Infrared said:
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
Sorry, I meant for that to be ##A_n=[1/2,1]\cup (1/(n+1),1/n)##
 
  • #27
Do me a favor and see if this is true, the Lerch Transcendent identity from my paper, for ##N\in\mathbb{Z}^+##, and I forget the domain of z and y, here it goes

$$\Phi (z,N,y) :=\sum_{q=0}^{\infty}\frac{z^q}{(q+y)^N}$$
$$=\int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}\prod_{k=1}^{N}\left( \lambda_k^{y-1}\right)\left( 1-z\prod_{q=1}^{N}\lambda_q\right)^{-1}\, d\lambda_1d\lambda_2\cdots d\lambda_N$$

Some of the results I got using that false theorem were known results that were actually true, but I didn't check all of them, if this identity holds then I still have something to work with, otherwise my paper is total crap, damn. I wrote this paper in junior college without any analysis classes under my belt, so I guess it is not a surprise it turned out this way. Thank you both for your help :)
 
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  • #28
How to handle the infinite discontinuity of the integrand at ##z=\lambda_k=1## for ##k=1,2,\ldots, N##? Do I take the upper bound of each integral to be ##1-\epsilon## and let ##\epsilon\rightarrow 0+##? Or do I have to set each upper bound to be ##1-\epsilon_k## and take a N-dimensional limit? Unsure
 
  • #29
I think there is either an alternate definition of limit of a sequence of sets that will make this theorem true or tighter hypotheses that make it true because I've verified all of my results by other means that I arrived at using this theorem. Perhaps connectedness is required? Perhaps compactness is required? Either of those would get rid of your counter-examples and still uphold my results. Maybe I'm barking up the wrong tree, it is possible that all those limits magically equaled what they were supposed to. I'm a novice mathematician here, should I go down this road or give up on it? Here's the thread where I verify by other means one of my results, check it out nobody in that forum has commented whether my work is correct or not, but I think it is. Thanks for your time and help.

Edit: I'd like to check whether the definition(s) of limit of sequence of sets set forth in this Wikipedia article will work for this theorem, it/they are different than Rudin's topological definition. A post will follow.
 
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  • #30
There's a few definitions of limit of a sequence of sets in the Wikipedia article I linked in the edit to the above post. The first of which is

Definition 1)
$$\limsup_{n\to\infty}A_n:=\bigcap_{n=1}^{\infty}\bigcup_{j=n}^{\infty}A_{j}$$
and
$$\liminf_{n\to\infty}A_n:=\bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}A_{j}$$

Then the definition of a limit of a sequence of sets: ##A_n\to A## if, and only if ##\exists## a set A such that ##\lim_{n\to\infty} A_{n} = \liminf_{n\to\infty}A_{n}=\limsup_{n\to\infty}A_{n}=A##

For my results to hold I require that

$$S_{n}^{N}:=\left\{\vec x \in\mathbb{R}^N | x_{k}\geq 0\forall k\, , \sum_{k=1}^{N}x_{k}^{2n}\leq N\right\}$$

converge to

$$S^{N} := \left\{ \vec x \in\mathbb{R}^N | 0 \leq x_{k} \leq 1 \forall k \right\}$$

Hence we require that

$$S_{+}^N:=\limsup_{n\to\infty}S_n^N:=\bigcap_{n=1}^{\infty}\bigcup_{j=n}^{\infty}S_{j}^N=S^N$$
and that
$$S_{-}^N:=\liminf_{n\to\infty}S_n^N:=\bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}S_{j}^N=S^N$$
so we need

$$S_{+}^N=\bigcap_{n=1}^{\infty}\bigcup_{j=n}^{\infty}\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\sum_{k=1}^{N}x_{k}^{2j}\leq N\right\} =\left\{ \vec x \in\mathbb{R}^N | 0 \leq x_{k} \leq 1 \forall k \right\}$$

and

$$S_{-}^N=\bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\sum_{k=1}^{N}x_{k}^{2j}\leq N\right\}=\left\{ \vec x \in\mathbb{R}^N | 0 \leq x_{k} \leq 1 \forall k \right\}$$

Edit: I need a break from writing this post, I'm going to post what I got now and edit later today... or one of you can prove the last two equalities please?

Here's a plot of ##\left\{x^{2*5}+y^{2*5}\leq 2,x^{2*25}+y^{2*25}\leq 2,x\geq 0, y\geq 0, x+y=2, y=x\right\}## the curved curve with a tighter corner is the curve ##x^{2*25}+y^{2*25}\leq 2## (yellow area)

implicitplot{x^{10}+y^{10} leq 2, x^{50}+y^{50} leq 2, y=x, x+y=2}.gif


so clearly we have ##S_{j+k}^N\subset S_{j}^N\, \, \forall j,k\geq 1## hence

$$S_{+}^N=\lim_{M\to\infty}\bigcap_{n=1}^{M}\bigcup_{j=n}^{M}\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\sum_{k=1}^{N}x_{k}^{2j}\leq N\right\} $$
$$S_{+}^N=\lim_{M\to\infty}\bigcap_{n=1}^{M}\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\sum_{k=1}^{N}x_{k}^{2n}\leq N\right\} $$
$$S_{+}^N=\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\lim_{M\to\infty}\sum_{k=1}^{N}x_{k}^{2M}\leq N\right\} $$
$$= \left\{ \vec x \in\mathbb{R}^N | 0 \leq x_{k} \leq 1 \forall k \right\}$$

and

$$S_{-}^N=\lim_{M\to\infty}\bigcup_{n=1}^{M}\bigcap_{j=n}^{M}\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\sum_{k=1}^{N}x_{k}^{2j}\leq N\right\}$$
$$=\lim_{M\to\infty}\bigcup_{n=1}^{M}\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\sum_{k=1}^{N}x_{k}^{2M}\leq N\right\} $$
$$=\left\{ \vec x \in \mathbb{R}^N | 0\leq x_k \forall k,\lim_{M\to\infty}\sum_{k=1}^{N}x_{k}^{2M}\leq N\right\} $$
$$= \left\{ \vec x \in\mathbb{R}^N | 0 \leq x_{k} \leq 1 \forall k \right\}$$

Hence ##S_n^N\to S^N## by this definition 1 of convergence of a sequence of sets. Check!
 
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  • #31
I found a problem on another site that solves my theorem troubles nicely.

Theorem 2) Let ##A_n , A\in\mathbb{R}^N## such that

$$A\subset\cdots\subset A_{n+1}\subset A_n\subset\cdots\subset A_2\subset A_1$$

and let ##A:=\bigcap_{n=1}^\infty A_n## Then for Lebesgue measurable ##f:\mathbb{R}^N\to\mathbb{C}## define the set function ##\phi : A_1\to\mathbb{C}## by
$$\phi (E):=\int_{E} f\, d\mu \, \, \forall E\subset A_1$$
Then,
$$\lim_{n\to\infty}\phi (A_n) = \phi (A)$$
which is to say explicitly that
$$\lim_{n\to\infty}\int_{A_n} f\, d\mu = \int_{A} f\, d\mu$$

Proof: Let
$$f_n:=f\cdot\chi_{A_n}=\begin{cases} f( x ) & \text{if } x \in A_n \\
0 & \text{ } \text{elsewhere} \end{cases}$$
and let
$$f:=f\cdot\chi_{A}=\begin{cases} f( x ) & \text{if } x \in A \\
0 & \text{ } \text{elsewhere} \end{cases}$$

Then
$$f_1 (x)\geq f_2 (x)\geq\cdots\geq f_n (x)\geq f_{n+1} (x)\geq\cdots f(x)$$
hence by the Lebesgue Dominated Convergence Theorem with ##f_1 (x) \geq | f_n (x) | \, \forall x\in \mathbb{R}^N,\forall n\in\mathbb{Z}^+## we have
$$\lim_{n\to\infty}\phi (A_n) = \lim_{n\to\infty}\int_{A_n} f\, d\mu = \int_{A} f\, d\mu =\phi (A) $$
and the theorem is demonstrated.

For my results to hold I require that I can use this sequence in Theorem 2:
$$S_{n}:=\left\{\vec x \in\mathbb{R}^N | x_{k}\geq 0\forall k\, , \sum_{k=1}^{N}x_{k}^{2n}\leq N\right\}$$
$$S :=\bigcap_{k=1}^\infty S_k = \left\{ \vec x \in\mathbb{R}^N | 0 \leq x_{k} \leq 1 \forall k \right\}$$

I'm not sure of how to prove via inequalities that
$$S\subset\cdots\subset S_{n+1}\subset S_n\subset\cdots\subset S_2\subset S_1$$
because in practice I used ##N=2## and lots of graphs to demonstrate the above property of ##S_{n+1}\subset S_n\, \forall n\in\mathbb{Z}^+##, any help would be appreciated. Thanks for your time!
 
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  • #32
You used the following conclusions:
\begin{align*}
\lim_{n \to \infty}\phi(A_n)&=\lim_{n \to \infty} \int_{A_n}f\,d\mu \\& \stackrel{(1)}{=} \lim_{n \to \infty} \int_{A}f\circ \chi(A_n)\,d\mu \\
&\stackrel{(2)}{=} \int_A \left(\lim_{n \to \infty} f\circ \chi(A_n) \right)\,d\mu \\
&\stackrel{(3)}{=} \int_A \left(f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right) \right) \,d\mu \\
&\stackrel{(4)}{=} \int_A f \circ \chi(A) \,d\mu \\
&=\int_A f\,d\mu\\
&= \phi(A)
\end{align*}
What is your argument for ##(3)##:
$$
\lim_{n \to \infty} f\circ \chi(A_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right)
$$
 
  • #33
I see a few of these different, so I'll change them in the quote:

fresh_42 said:
You used the following conclusions:
\begin{align*}
\lim_{n \to \infty}\phi(A_n)&=\lim_{n \to \infty} \int_{A_n}f\,d\mu \\& \stackrel{(1)}{=} \lim_{n \to \infty} \int_{\mathbb{R}^N}f\circ \chi(A_n)\,d\mu \\
&\stackrel{(2)}{=} \int_{\mathbb{R}^N} \left(\lim_{n \to \infty} f\circ \chi(A_n) \right)\,d\mu \\
&\stackrel{(3)}{=} \int_{\mathbb{R}^N} \left(f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right) \right) \,d\mu \\
&\stackrel{(4)}{=} \int_A f \circ \chi(A) \,d\mu \\
&=\int_A f\,d\mu\\
&= \phi(A)
\end{align*}
What is your argument for ##(3)##:
$$
\lim_{n \to \infty} f\circ \chi(A_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right)
$$

Since the ##A_n## are decreasing we have that ##f\circ \chi (A_n )\downarrow f\circ\chi (A)## where the convergence is point-wise and because every point of ##A## is in every ##A_n## and so by Dominated Convergence Theorem with ##\left| f\circ \chi (A_n )\right| \leq \left| f\circ\chi (A_1)\right|\in\mathfrak{L}(\mu )## because ##f\in\mathfrak{L}(\mu )## and I think we can assume the ##A_n## are measurable subsets of ##\mathbb{R}^N##, right? So by DCT we have
$$\lim_{n\to\infty}\int_{\mathbb{R}^N} f\circ\chi (A_n ) \, d\mu = \int_{\mathbb{R}^N}\lim_{n\to\infty}f\circ\chi (A_n )\, d\mu = \int_{\mathbb{R}^N} f\circ\chi (A)\, d\mu = \int_A f\, d\mu$$

Edit: There was also a note about using Monotone Convergence Theorem instead of DCT.
 
  • #34
I'm still not sure whether ##(3)## doesn't need continuity or boundness of ##f##. You pull the limit across the function, so why is this allowed?

I'm not saying it's wrong, only that I don't see it. However, I'm not quite sure how strong Lebesgue integrable is.
 
Last edited:
  • #35
Think of the definition of the infinite limit

$$\lim_{n\to\infty}f\circ\chi \left( \bigcap_{k=1}^{n} A_k \right) = f\circ\chi (A)\Leftrightarrow $$
$$\forall \epsilon >0\, \exists\, M\in\mathbb{Z}^+\text{ such that } n\geq M\Rightarrow \left|f\circ\chi \left( \bigcap_{k=1}^{n} A_k \right)- f\circ\chi (A) \right|<\epsilon$$

Here we are always dealing finite number of intersections and we know that

$$A\subset \cdots \subset A_{n+1}\subset A_{n}\subset\cdots \subset A_{2}\subset A_1$$

hence ascertain that ##\bigcap_{k=1}^{n} A_k = A_n## and inserting this into the above definition of a limit we obtain

$$\forall \epsilon >0\, \exists\, M\in\mathbb{Z}^+\text{ such that } n\geq M\Rightarrow \left|f\circ\chi (A_n) - f\circ\chi (A) \right|<\epsilon$$
$$\Leftrightarrow\lim_{n\to\infty}f\circ\chi (A_n) = f\circ\chi (A) $$

Does that answer your question?
 

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