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The unit disc contains finitely many dyadic squares whose total area is arbitrarily close to the area of the disc

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Exercise 1.25.a in Pugh’s Real Mathematical Analysis states that

Given $\epsilon > 0$, show that the unit disc contains finitely many dyadic squares whose total area exceeds $\pi - \epsilon$, and which intersect each other only along their boundaries.

My proof goes as follows. Let $B$ be the unit disc and $C$ be a smaller concentric disc of an arbitrary radius, say $1-\delta$ for a positive $\delta$. We will cover $C$ with finitely many dyadic squares in such a way that the squares completely lie inside $B$.

Consider the first quadrant. Let $k$ be so large that $1/2^k\lt \delta$. Start with the square $[0,1/2^k]^2$ and extend it to the right until the rightmost square hits $C$. If it hit $B$ too then consider a smaller $k’$ such that the new rightmost square hits $C$ but not $B$, and then fill the previous squares with the new ones of smaller size, paying attention to how to fill the the previous rightmost square in an appropriate way which will allow us to discard the smaller squares which hit $B$.

Now go one step up with the same squares of size $1/2^k$ and fill them appropriately with the smaller squares of size $1/2^{k’}$. If the rightmost squares (lying in different heights) hit $B$, choose a smaller $k’’$ and then fill all the squares in this and the previous step with the new squares of size $1/2^{k’’}$.

Continue this procedure until you reach to the top of $B$. As $k$ is a finite number, the number of steps needed to complete the procedure is finite, so there is a finite number $k^*$ such that finitely many dyadic squares of size $1/2^{k^*}$ cover $C$ and at the same time lie inside $B$.

As $\delta$ was arbitrary, we are done.

Is my proof correct? Is there any simpler solution?


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