# Convergence of a sequence of sets

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• benorin
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.
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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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:

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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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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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.

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...

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.

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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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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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.)

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.

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.##

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##.

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.##

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)?##

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.

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##.

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##.

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.

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.

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.

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.

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.

No, I gave counterexamples in posts 9 and 11

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)...

Sorry, I meant for that to be ##A_n=[1/2,1]\cup (1/(n+1),1/n)##

Well holy cow, it's not true!

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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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

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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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)

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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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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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)##:
$$\lim_{n \to \infty} f\circ \chi(A_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right)$$

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.

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:
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

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

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