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:


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

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

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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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