I am studying Measure Theory from these notes. The author passes off the fact that the function $(x,r)\mapsto A_r f(x)$ is continuous where $f$ is locally integrable function as something easy. Here's the context:
Here's my attempt at proving this: Let $f\in L^{1}_{loc} \left( m_{n} \right)$. Consider the map $F: \mathbb R^{n} \times (0, \infty) \to \mathbb R$ given by $F(x,r)=A_r f(x)$ for each $(x,r) \in \mathbb R ^{n} \times (0, \infty)$. We need to show that this map is continuous.
Let $(x,r)\in \mathbb R^{n} \times (0, \infty)$ and let $\left( x_{n}, r_{n} \right)$ be a sequence in $\mathbb R ^{n} \times (0,\infty)$ converging to $(x,r)$. We wish to show that $A_{r_n} f(x_n) \to A_r f(x)$.
We first show that $\int_{B(x_{n}, r_{n})} f(y) dy \to \int_{B(x,r)} f(y) dy$ (Call it goal $(\star)$). Since the sequence $\left( x_n, r_{n} \right)$ converges, it must be bounded. Therefore, there exists a constant $K>0$ such that $\lVert(x_n, r_n)\rVert_{2}<K$ for each $n \in \mathbb N$. Hence, we have that $(x_{n}, r_{n}) \in B[0, K]$ for each $n \in \mathbb N$. Now, note that $|f\chi_{B(x_{n}, r_{n})}| \le |f\chi_{B[0,K]}|$ for each $n\in \mathbb N$ and $f\chi_{B[0,K]}$ is in $L^{1} \left( m_{n} \right)$ by our assumption that it is locally integrable.
We now show that $\chi_{B(x_{n}, r_n)} \to \chi_{B(x,r)}$ pointwise. Let $y \in B(x,r)$. Then we have that $\chi_{B(x,r)}(y)=1$ and hence we need to show that $\chi_{B\left( x_{n},r_{n} \right)} =1$ eventually. To see this, consider the sequences $|y-x_{n}|$ and $r_n$. Since $|y-x_{n}| \to |y-x|$, $r_n \to r$ and $|y-x|<r$, we have that $|y-x_n|<r_n$ eventually. This shows that $\chi_{B(x_{n}, r_{n})}(y)=1$ eventually. A similar argument shows that if $y\not\in B(x,r)$ then $y\not\in B(x_{n},r_n)$ eventually.
Hence, we have that $f\chi_{B(x_{n},r_{n})} \to f\chi_{B(x,r)}$ pointwise. Since we meet all the requirements to apply the dominated convergence, we do so to conclude $(\star)$.
Since $r_{n} \to r$, we have that $\limsup_{n\in \mathbb N} B(0,r_{n})=\liminf_{n\in \mathbb N} B(0,r_n) = B(0,r)$. Hence, we have by continuity of measures that $\lim_{k\in \mathbb N} m_{n} \left( B(0,r_{k}) \right) = m_{n} \left( B(0,r) \right)$. Also, since Lebesgue measure is translation invariant, we have that $\lim_{k\in \mathbb N} m_{n} \left( B(x_k,r_{k}) \right) = m_{n} \left( B(x,r) \right)$.
Using these two facts, we have shown that $(x,r) \mapsto A_{r} f(x)$ is continuous.
Although I think my argument is correct, I feel there must be easier way to prove this else the author would not have skipped the proof. I would appreciate it if people can offer an alternative and shorter proof of this.
What you did is almost correct: the convergence $\chi_{B(x_k,r_k)}\to \chi_{B(x,r)}$ may fail pointwise: for example, if $n=1$, $x=x_k=0$ and $r_k=1+(-1)^k/k$, we do not have the convergence for $x=\pm 1$. However, the only place where the convergence $\chi_{B(x_k,r_k)}\to \chi_{B(x,r)}$ may fail is at points $y$ such that $\lvert y-x\vert=r$ and this set has Lebesgue measure zero.
A minor point: since $n$ is used for the dimension, it would be more safe to write $(x_k,r_k)$ instead of $(x_n,r_n)$.