This question comes from problem #3 of the Spring 2013 Analysis Qual Exam here http://www.math.ucla.edu/grad/handbook/hbquals.shtml
Define for $f\in C(\mathbb{R}^{2}\to\mathbb{R})$ $$(A_{r}f)(x,y):=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+r\cos\theta,y+r\sin\theta)\;d\theta$$ and $$(Mf)(x,y)=\sup_{0<r<1}(A_{r}f)(x,y).$$ Assuming a theorem of Bourgain that there exists some absolute constant $C$ such that $Mf$ is bounded on $C_{c}(\mathbb{R}^{2}\to\mathbb{R})$, i.e. $$||Mf||_{L^{3}(\mathbb{R}^{2})}\leq C||f||_{L^{3}(\mathbb{R}^{2})},$$ prove that if $K\subset\mathbb{R}^{2}$ is compact then one has $$\lim_{r\to0}\;(A_{r}\chi_{K})(x,y)=1\;\text{a.e.}\;x.$$ $\chi_{K}$ is the characteristic function of $K$.
(Proposed) Proof. If $f$ is continuous on $\mathbb{R}^{2}$ then so is each $g_{r}(\theta):=f(x+r\cos\theta,y+r\sin\theta)$ for each fixed $(x,y)$. With $(x,y)$ fixed and $r<1$ (say), $$\sup_{0<r<1}||g_{r}||_{L^{\infty}}\leq\sup_{0<r<1}\sup_{\partial B(r;(x,y))}|f(s,t)|=\sup_{B(1;(x,y))}|f(s,t)|=M(x,y)<\infty.$$ And so $\{g_{r}(\theta)\}_{0<r<1}$ is dominated by $M(x,y)$, which is integrable on $[-\pi,\pi]$. Thus by continuity of the integrands and the dominted convergence theorem, \begin{align*} \lim_{r\to0}\;(A_{r}f)(x,y) &=\frac{1}{2\pi}\int_{-\pi}^{\pi}\lim_{r\to0}g_{r}(\theta)\;d\theta \\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\lim_{r\to0}(x+r\cos\theta),\lim_{r\to0}(y+r\sin\theta))\;d\theta\\ &=f(x,y)\;\text{a.e.}\;x.\end{align*}
In the special case that $f=\chi_{K}$, we see that $f(x,y)=1$ for $(x,y)\in K$ and $f(x,y)=0$ for $(x,y)\in K^{c}$. Also, the monotone convergence theorem can be used instead in this case since with $\chi_{K}$ as our definition for the $g_{r}$, as $r\to0$, the $g_{r}$ form an increasing sequence of truncations of $\chi_{K}$ (as a function of $\theta$).
Obviously if someone could point out the flaw in my reasoning, I'd much appreciate it. In particular, why do I need to use the maximal function, and in particular the boundedness of it on $L^{3}\cap C_{c}\mapsto L^{3}$. Also, why is there a special interest in $\chi_{K}$ and $K$ being compact? These assumptions lead me to believe the proof is much more complicated for general continuous $f$ than I have indicated in my attempted proof.
Edit 1. When passing the limit $\lim_{r\to0}$ under the argument of the integrand, I assumed continuity of $f$. But $\chi_{K}$ isn't necessarily continuous on $B(1,(x,y))$, hence $g_{r}(\theta)$ neednot be continuous on $[-\pi,\pi]$ for any $0<r<1$ and so passing the limit into the argument cannot be (immediately) justified. If $K$ contains a ball about $(x,y)$, then for $r$ sufficiently small we are fine, but $K$ may be sufficiently "bad" to the point where there is no immediate justification (actually, is this possible for compact sets? We needed to only prove the assertion for a.e. $x$ so we could in principle ignore points on the boundary of $K$ or isolated points). This must be where more sophisticated arguments employing the maximal function are required? I suspect some kind of approximation argument with continuous functions with compact support being dense in $L^{3}$, but I am not sure how $L^{3}$ enters into the problem...
