Verifying pointwise convergence of indicator functions

[psie]

TL;DR Summary

I'm stuck at something fairly basic I think. Let $B(r,x)$ be an open ball of radius $r$ and center $x$ in $\mathbb R^n$. It is claimed that $\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}$ pointwise as $r\to r_0$ and $x\to x_0$ on $\mathbb R^n\setminus S(r_0,x_0)$, where $S(r_0,x_0)$ is the sphere $\{y:|y-x_0|=r_0\}$. I am stuck showing this.

I'm reading a proof of a lemma that $$A_rf(x)=\frac1{m(B(r,x))}\int_{B(r,x)}f(y)\,dy,$$where $m$ is Lebesgue measure, is jointly continuous in $r$ and $x$ ($A$ stands for average). The claim that $\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}$ on $\mathbb R^n\setminus S(r_0,x_0)$ is made in the proof. I think there are two cases to consider. Let $y\in \mathbb R^n\setminus S(r_0,x_0)$.

1. $|y-x_0|<r_0$, i.e. $\chi_{B(r_0,x_0)}(y)=1$. Is it then also true that for some sequences $(x_n),(r_n)$ that converge to $x_0,r_0$ respectively, that $|y-x_n|<r_n$ for large enough $n$? Why? If yes, then $\chi_{B(r_n,x_n)}(y)=1$ for large enough $n$ too.
2. Similarly, if $|y-x_0|>r_0$, is it then true that $|y-x_n|>r_n$?

Also, why does $\chi_{B(r,x)}\not\to\chi_{B(r_0,x_0)}$ on the sphere $S(r_0,x_0)$?

 

[fresh_42]

My intuition says 1) yes, 2) possibly, and 3) because you cannot avoid the jump from $0$ to $1$ on the sphere as there is no neighborhood completely contained in either region. However, I have to consider the triangle inequalities in detail. I think the answer to 3) is the key to the first two questions.

 

[psie]

My intuition says 1) yes, 2) possibly, and 3) because you cannot avoid the jump from $0$ to $1$ on the sphere as there is no neighborhood completely contained in either region. However, I have to consider the triangle inequalities in detail. I think the answer to 3) is the key to the first two questions.

Ok. I agree, I don't see how to derive $|y-x_n|<r_n$ from the triangle inequality given $|y-x_0|<r_0$ and $x_n\to x_0,r_n\to r_0$. It doesn't lead anywhere: $$|y-x_n|\leq |y-x_0|+|x_0-x_n|<r_0+\epsilon.$$Here $|x_0-x_n|<\epsilon$ for sufficiently large $n$. But I don't see how else to show $$\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}$$pointwise on $\mathbb R^n\setminus S(r_0,x_0)$.

 

[psie]

Actually, I worked out 1).

Let $\epsilon= r_0 - |y - x_0|$. For sufficiently large $n$, $|x_n - x_0| < \epsilon/2$ and $|r_n - r_0| < \epsilon/2 \implies r_n > r_0 - \epsilon/2$. So $|y - x_n| < (r_0 - \epsilon) + \epsilon/2 < r_n$.

 

[fresh_42]

Ok. I agree, I don't see how to derive $|y-x_n|<r_n$ from the triangle inequality given $|y-x_0|<r_0$ and $x_n\to x_0,r_n\to r_0$. It doesn't lead anywhere: $$|y-x_n|\leq |y-x_0|+|x_0-x_n|<r_0+\epsilon.$$Here $|x_0-x_n|<\epsilon$ for sufficiently large $n$. But I don't see how else to show $$\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}$$pointwise on $\mathbb R^n\setminus S(r_0,x_0)$.

We may assume w.l.o.g. that $x_0=0$ and $r_0=1.$ We may also assume that $x_n=x_1/n$ where $x_1=1-c$ for some constant $c\in (0,1).$ We then need the condition $B(x_n,r_n)\stackrel{(*)}{\subset} B(0,1).$ I struggle a bit to define $r_n$ appropriately such that $\lim_{n \to \infty}r_n=1$ and $(*)$ still holds, but that would be the idea.

The second is analogous with $B(x_n,r_n)\subseteq \mathbb{R}^n\setminus B(0,1).$