Convergence of a sequence of sets

  • #26
benorin
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Well holy cow, it's not true!
 
  • #27
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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
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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
 
  • #29
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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
benorin
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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 gonna 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
benorin
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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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  • #32
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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)##:
$$
\lim_{n \to \infty} f\circ \chi(A_n) {=}_{a.e.} f \circ \chi\left(\displaystyle{\bigcap_{n=1}^{\infty}}A_n\right)
$$
 
  • #33
benorin
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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.
 
  • #34
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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.
 
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  • #35
benorin
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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?
 
  • #36
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No. My problem was exactly why ##\|f(\chi(A_n))(x)-f(\chi(A))(x)\|<\varepsilon## without further conditions on ##f##.
 
  • #37
benorin
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Well we know that f is integrable and ##A_n## is measurable, is that not enough?
 
  • #38
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Well we know that f is integrable and ##A_n## is measurable, is that not enough?
I am not sure. I suppose we need ##f## to be continuous, but maybe Lebesgue integrable will do. I just don't see it. We have that ##\|\chi(A_n)(x)-\chi(A)(x)\|## is small, but why does applying ##f## keep this condition?
 
  • #39
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I found a remark in Baby Rudin that I think may help,

Remark: If f is Lebesgue integrable on E, and if ##A\in\mathfrak{M}## (A is measurable) and ##A\subset E##, then f is Lebesgue integrable on A.

Also that if f is Lebesgue integrable on E, the f is finite on E a.e. may also help, I think
 
  • #40
benorin
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What if I let ##B_n:=\bigcap_{k=1}^n A_k## and then define ##f_k:=f\circ\chi (B_k)## and proceed with the proof as before, would

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

still give you pause?
 
  • #41
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Sure, but what makes you confident, that ##f## doesn't destroy that behavior? If ##f## is continuous, o.k., but is it?
 
  • #42
benorin
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Let ##B_n:=\bigcap_{k=1}^n A_k## and then define ##f_k:=f\circ\chi (B_k)## and proceed with the proof as before, let's examine whether

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

is true: is the measure of the set where they're not equal equal to zero? I.e. is

$$\mu \left( S\left( \lim_{n\to\infty}B_n , \bigcap_{k=1}^{\infty}A_k\right) \right)$$
$$=\mu \left\{\left( \left( \lim_{n\to\infty}B_n\right) - \bigcap_{k=1}^{\infty}A_k\right) \cup \left(\left(\bigcap_{k=1}^{\infty}A_k\right) - \left( \lim_{n\to\infty}B_n\right)\right) \right\}=0?$$

where ##S(A,B)## is the symmetric difference of sets; I think this follows from the definnition of ##B_n## trivially. So, satisfied that the ##\chi 's## are equal everywhere it follows that ##f\circ \chi 's## are also equal since applying f just inserts the value of f(x) if x is in the set ##\chi## is of (and zero otherwise) and these are equal sets on both sides so evaluating f in the same places on both sides should be equal everywhere, right?
 
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  • #43
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"Let S,Sn be as in the problem statement. It is clear that S is the (increasing) union of the Sn. This is already enough to guarantee that m(S(S,Sn))=m(S−Sn) goes to zero as n→∞."

To be airtight, it's necessary to account for the boundary ∂S of S.
 
  • #44
Infrared
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Why? I gave an argument- do you disagree with any part of it?
 
  • #45
benorin
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@zinq , I'm uncertain which post you are quoting but I should make you aware there has been at least 3 different definitions of convergence of a sequence of sets employed at various times in this thread. I ended up using point-wise convergence and the proof of post #31. As for showing that the particular sequence of sets I mentioned I needed to work with the theorem in post #31 I have used bivariate induction and three different Lagrange Multipliers proofs to flesh out the induction proof (not shown here). If you mean to point out a shortcoming of the proof in post #31, then I am most interested in what you have to say.
 

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