3
$\begingroup$

Let $A,B\in\mathcal A_{\Bbb R}^*$ given with $\overline{\lambda}(A)<\infty$ and $\overline{\lambda}(B)<\infty$. Lets define $\; \overline{\lambda}_{A,B}:\Bbb R\to\Bbb R$ as follows:

$$\overline{\lambda}_{A,B}(x)=\overline{\lambda}(A\cap(B+x))$$

where $B+x=\{b+x:b\in B\}$.

So what I want to prove is that $\overline{\lambda}_{A,B}$ is continuous.


Proof:
Let $c\in\Bbb R$ fixed and let $x_n=c-\frac{1}{n}\;\forall n\in\Bbb N$. Clearly $x_n\in\Bbb R\;\forall n\in\Bbb N$.

Then, let $B_n=B+x_n\;\forall n\in\Bbb N\Rightarrow\ B_n=\{b+c-\frac{1}{n}:b\in B\}\;\forall n\in\Bbb N$


Lemma 1: $$B_n\subseteq B_{n+1}\;\forall n\in\Bbb N$$
Let $y\in B_n\Rightarrow\ y=b+c-\frac{1}{n}=b+c-\big(\frac{1}{n(n+1)}+\frac{1}{n+1}\big)=b+c-\frac{1}{n(n+1)}-\frac{1}{n+1}\;\forall n\in\Bbb N$. Thus $ y=b+c'-\frac{1}{n+1}\;\forall n\in\Bbb N$ with $c'=c-\frac{1}{n(n+1)}\Rightarrow\ y\in B_{n+1}$.


Lemma 2: $$\lim_{n\to\infty}(A\cap B_n)=A\cap (B+c)$$
Since $B_n\subseteq B_{n+1}\;\forall n\in\Bbb N$ by Lemma 1, the limit of $(B_n)_{n\in\Bbb N}$ exists and $\lim_{n\to\infty}(B_n)=\bigcup_{n=1}^\infty B_n,\ $but $\bigcup_{n=1}^\infty B_n=\bigcup_{n=1}^\infty \{b+c-\frac{1}{n}:b\in B\}=\{b+c:b\in B\}=B+c$. Now, clearly $A\cap B_n\subseteq A\cap B_{n+1}\;\forall n\in\Bbb N\Rightarrow\ \lim_{n\to\infty}(A\cap B_n)=\bigcup_{n=1}^\infty (A\cap B_n)=A\cap\bigcup_{n=1}^\infty B_n=A\cap (B+c)$.


So by Lemma 2 we get that: $$\lim_{n\to\infty}\overline{\lambda}_{A,B}(x_n)=\lim_{n\to\infty}\overline{\lambda}(A\cap(B+x_n))=\lim_{n\to\infty}\overline{\lambda}(A\cap B_n)=\overline{\lambda}\big(\lim_{n\to\infty}(A\cap B_n)\big)=\overline{\lambda}(A\cap (B+c))=\overline{\lambda}_{A,B}(c)$$

And clearly $\lim_{n\to\infty}x_n=c$, thus $\overline{\lambda}_{A,B}$ is continuous.

Did I miss something?

$\endgroup$
  • $\begingroup$ I get the impression you only show $\bar{\lambda}_{A,B}(c-\frac{1}{n})\to \bar{\lambda}_{A,B}(c)$. I think you have to show, for any sequence $x_n\to c$, $\bar{\lambda}_{A,B}(x_n)\to \bar{\lambda}_{A,B}(c)$. Also, in the proof of lemma 1, why should $b+c'-\frac{1}{n+1}$ belong to $B_{n+1}$? $\endgroup$ – Nate River Nov 30 '15 at 11:19
  • $\begingroup$ Yes, miss that, thanks. I trying to do it for any sequence $x_n\to c$, but can't managed to get anything, any idea? And since $B_{n+1}=\{b+c-\frac{1}{n+1}: b\in B\}$, $b+c'-\frac{1}{n+1}$ is an element. $\endgroup$ – Arnulf Nov 30 '15 at 14:28
  • $\begingroup$ I don't think $b+c'-\frac{1}{n+1}$ is an element of $B_{n+1}$. Consider $B=\{0\}$, $c=0$. Then $B_{n+1}=\{-\frac{1}{n+1}\}$. Also, $c'=-\frac{1}{n(n+1)}$, so $c'-\frac{1}{n+1}=-\frac{1}{n}\notin B_{n+1}$. $\endgroup$ – Nate River Nov 30 '15 at 15:07
  • $\begingroup$ Oh I see it now, thanks. So what do you suggest for proving this function is continuos? $\endgroup$ – Arnulf Nov 30 '15 at 15:12
  • 1
    $\begingroup$ As for ideas for the proof. Are you familiar with the dominated convergence theorem? In case you do, a possible hint is this: Consider the functions $1_{A\cap (B+x_n)}$ and the function $1_{A\cap(B+c)}$. Do you have pointwise convergence almost everywhere? Also, find an integrable function which dominates the functions $1_{A\cap (B+x_n)}$. $\endgroup$ – Nate River Nov 30 '15 at 15:13
1
$\begingroup$

So continuing with Nate River's hint, let $(x_n)_{n\in\Bbb N}$ be any sequence in $\Bbb R$ s.t. $\lim_{n\to\infty} x_n=c\;\ \forall c\in\Bbb R$ fixed.

Then, let $f_n=\chi_{A\cap (B+x_n)}\;\ \forall n\in\Bbb N\Rightarrow\ (f_n)_{n\in\Bbb N}\subset M(\Bbb R,\mathcal A_{\Bbb R}^*)\;\;\forall n\in\Bbb N$. And let $f=\chi_{A\cap (B+c)}\in M(\Bbb R,\mathcal A_{\Bbb R}^*)$


Lemma 1: $f_n\to f$ as $n\to\infty$
Let $E_n=A\cap (B+x_n)\;\;\forall n\in\Bbb N$. So, clearly $\; \underline{\lim}_{n\to\infty}(E_n)\subseteq \overline{\lim}_{n\to\infty}(E_n)$. Then, if $x\in\overline{\lim}_{n\to\infty}(E_n)$ we know that:

$$\Rightarrow\ x\in E_n\;\;\text{for infite values of $n$}\\ \Rightarrow\ x\in A\cap (B+x_n)\;\;\text{for infite values of $n$}\\ \Rightarrow\ x\in B+x_n\;\;\text{for infite values of $n$} \text{ and } x\in A\\ \Rightarrow\ \exists\ b\in B,\;\ x=b+x_n\;\;\text{for infite values of $n$} \text{ and } x\in A\\ \Rightarrow\ \exists\ b\in B,\;\ x=\lim_{n\to\infty}b+x_n=b+c \text{ and } x\in A\\ \Rightarrow\ x\in B+c \text{ and } x\in A$$

thus $\overline{\lim}_{n\to\infty}(E_n)\subseteq A\cap (B+c)$.

Now, if $x\in A\cap (B+c)$ we know that:

$$\Rightarrow\ \exists\ b\in B,\;\ x=b+c \text{ and } x\in A$$

and, since $\lim_{n\to\infty} x_n=c,$ given $\epsilon>0\ \exists\ n_0\in\Bbb N$ s.t. $\forall n\ge n_0\;\; x=b+x_n$

$$\Rightarrow\ \exists\ b\in B,\;\ x=b+x_n\;\forall n\ge n_0\;\;\text{and}\; x\in A\\ \Rightarrow\ x\in B+x_n\;\;\forall n\ge n_0\;\;\text{and}\; x\in A\\ \Rightarrow\ x\in A\cap (B+x_n)\;\;\forall n\ge n_0\\ \Rightarrow\ x\in E_n\;\;\forall n\ge n_0\Rightarrow\ x\in\underline{\lim}_{n\to\infty}(E_n)$$

thus $A\cap (B+c)\subseteq \underline{\lim}_{n\to\infty}(E_n)$.

Thereby $\underline{\lim}_{n\to\infty}(E_n)=\lim_{n\to\infty}(E_n)=\overline{\lim}_{n\to\infty}(E_n)\Rightarrow\ (\chi_{E_n})_{n\in\Bbb N}$ converges, where:

$$A\cap (B+c)\subseteq \underline{\lim}_{n\to\infty}(E_n)=\lim_{n\to\infty}(E_n)=\overline{\lim}_{n\to\infty}(E_n)\subseteq A\cap (B+c)\\ \text{so}\;\; \lim_{n\to\infty}(E_n)=A\cap (B+c)$$

and thus $(\chi_{E_n})_{n\in\Bbb N}$ converges to $\chi_{\lim_{n\to\infty}(E_n)}=\chi_{A\cap (B+c)}$.

Thus $f_n=\chi_{A\cap(B+x_n)}\to\chi_{A\cap (B+c)}=f$ as $n\to\infty$


Now let $g=\chi_A$ so, since $A\cap (B+x_n)\subseteq A\;\;\forall n\in\Bbb N$, we get that $|f_n|\le g\;\;\forall n\in\Bbb N$ where: $\int gd\overline{\lambda}=\overline{\lambda}(A)<\infty\Rightarrow\ g\in\mathcal L_1(\overline{\lambda})$. So applying LDCT (Lebesgue Dominated Convergence Theorem) to $(f_n)_{n\in\Bbb N}$ we get that:

$$\lim_{n\to\infty}\int f_n d\overline{\lambda}=\int fd\overline{\lambda}\\ \Leftrightarrow\ \lim_{n\to\infty}\int \chi_{A\cap (B+x_n)} d\overline{\lambda}=\int \chi_{A\cap (B+c)}d\overline{\lambda}\\ \Leftrightarrow\ \lim_{n\to\infty} \overline{\lambda}(A\cap (B+x_n))=\overline{\lambda}(A\cap (B+c))\\ \Leftrightarrow\ \lim_{n\to\infty} \overline{\lambda}_{A,B}(x_n)=\overline{\lambda}_{A,B}(c)$$ thereby $\overline{\lambda}_{A,B}$ is continuous.

$\endgroup$
  • $\begingroup$ Sorry for not answering before. I was very busy. Anyway. The problem I see with this proof is that $\limsup A\cap(B+x_n)$ might be different from $\liminf A\cap(B+x_n)$. In particular, the inclusion $A\cap(B+c)\subset \liminf A\cap(B+x_n)$ might be false. Consider $A=[0,1],B=\{0\},c=0$ and $x_n=1/n$. What can you say about $\liminf A\cap(B+x_n)$? $\endgroup$ – Nate River Dec 4 '15 at 12:52
  • $\begingroup$ As for possible proofs. A possible way to prove it, is: first, prove it for sets of the form $A=(a,b], B=(c,d]$. Then for the semi-algebra of those sets. And finally extend the argument to general measurable sets. $\endgroup$ – Nate River Dec 4 '15 at 12:55
  • $\begingroup$ Thanks! Didn't noticed that little issue. I'm trying with your idea but got stuck. So $B+x_n=(c+x_n,d+x_n]$ and then: $$A\cap (B+x_n)= \left\{ \begin{array}{ll} \emptyset & b< c+x_n\;\;\text{or}\;\;d+x_n<a \\ (c+x_n,b] & a< c+x_n< b \\ (a,d+x_n] & c+x_n< a< d+x_n \\ A & A\subset (B+x_n) \\ B+x_n & (B+x_n)\subset A \end{array} \right. $$ I don't know If I'm doing it right. $\endgroup$ – Arnulf Dec 6 '15 at 21:15
  • $\begingroup$ Yes, you correctly computed the set $A\cap (B+x_n)$. The set $A\cap (B+c)$ also looks like this. If $x$ is in the interior of $A\cap (B+c)$, you can try to show $1_{A\cap(B+x_n)}(x) \to 1_{A\cap (B+c)}(x)$. Can you apply the dominated convergence theorem in this case? $\endgroup$ – Nate River Dec 7 '15 at 10:55
  • $\begingroup$ Ok, but I'm getting confused with the fact of showing $x$ is in $int[A\cap (B+c)]$. The procedure should be to take any $x\in A\cap (B+x_n)$ and try to find some $\delta>0$ s.t. $(x-\delta,x+\delta)\subset A\cap (B+c)$ ? $\endgroup$ – Arnulf Dec 7 '15 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.