Undergrad Convergence of a sequence of sets

[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?

 

[fresh_42]

No. My problem was exactly why $\|f(\chi(A_n))(x)-f(\chi(A))(x)\|<\varepsilon$ without further conditions on $f$.

 

[benorin]

Well we know that f is integrable and $A_n$ is measurable, is that not enough?

 

[fresh_42]

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?

 

[benorin]

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

 

[benorin]

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?

 

[fresh_42]

Sure, but what makes you confident, that $f$ doesn't destroy that behavior? If $f$ is continuous, o.k., but is it?

 

[benorin]

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?

 

[zinq]

"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.

 

[Infrared]

Why? I gave an argument- do you disagree with any part of it?

 

[benorin]

@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.