Convergence of a sequence of sets

[benorin]

TL;DR Summary

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!

 

[fresh_42]

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.

 

[benorin]

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 $S_n\rightarrow S$.

 

[fresh_42]

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.

 

[benorin]

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]

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.

 

[benorin]

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.

 

[benorin]

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?

 

[Infrared]

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.)

 

[benorin]

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.

 

[Infrared]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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.

 

[benorin]

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]

No, I gave counterexamples in posts 9 and 11

 

[benorin]

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 $A_n=[1/2,1]\cup (1/(n+1),1/n)$

 

[benorin]

Well holy cow, it's not true!

 

[benorin]

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 :)

 

[benorin]

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

 

[benorin]

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.

 

[benorin]

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!

 

[benorin]

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!

 

[fresh_42]

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 $(3)$:
$$
\lim_{n \to \infty} f\circ \chi(A_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right)
$$

 

[benorin]

I see a few of these different, so I'll change them in the quote:

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.

 

[fresh_42]

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.

 

[benorin]

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?