# Convergence of a sequence of sets

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Homework Helper

## 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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fresh_42
Mentor
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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Homework Helper
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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fresh_42
Mentor
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.

Homework Helper
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...

fresh_42
Mentor
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.

Homework Helper
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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Homework Helper
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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Infrared
Gold Member
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.)

Homework Helper
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.

Infrared
Gold Member
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 $f(x)=1/x^2.$

fresh_42
Mentor
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 $\mu$.

Infrared
Gold Member
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

$\lim_{n\to\infty}\int_{S_n}g dm=\lim_{n\to\infty}\int_{\mathbb{R}^p} g\chi_n dm=\int_{\mathbb{R}^p} \lim_{n\to\infty} g\chi_n dm=\int_{\mathbb{R}^p} g\chi dm=\int_{S}g dm.$

fresh_42
Mentor
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 $m(A)\leq m(B)?$

Infrared
Gold Member
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.

fresh_42
Mentor
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 $\mu \neq \lambda$.

Infrared
Gold Member
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 $0$.

fresh_42
Mentor
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.

Infrared
Gold Member
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.

fresh_42
Mentor
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.

Infrared
Gold Member
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.

Homework Helper
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.

Infrared
Gold Member
No, I gave counterexamples in posts 9 and 11

Homework Helper
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)...

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