Averages of Indicator Functions of Compact Sets in $\mathbb{R}^{2}$ Over Circles 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...
 A: First, if you just want to show that $A_r f(x,y) \to f(x,y)$ pointwise (a.e.) for continuous functions, there is no need to compare to the maximal function $M$  because you can simply use the fact that $f$ is bounded (and as I remarked, the bound may not even be true).
As your edit mentions, the characteristic function is not continuous so you cannot apply directly. And you are right to suspect an approximation argument. Note with just an approximation argument, you can already show that $A_r \chi_K \to \chi_K$ in the sense of $L^2$ (or any $L^p$, $1\leq p<\infty$, really, just need density).
The role of the maximal function in general is to show a.e. convergence, which is done in a similar manner to http://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem#Discussion_of_proof
Since you have an $L^3$ estimate, do your approximate $\chi_K$ with a continuous function of compact support in $L^3$, i.e. for any $\epsilon$ find $g_\epsilon \in C_c$ with $\|\chi_K - g_\epsilon \|_3 \leq \epsilon$.
Also use density to show that the maximal inequality bound holds for any $L^3$ function, i.e. $\|Mf\|_3 \leq C\|f\|_3$ for any $f\in L^3$. 
I think this should be enough for you to finish. If $h = \chi_K - g_\epsilon$, you would show that the measure of the set where $|h|>\alpha$ is small and the set where $|Mh| > \alpha$ is also small.
