There is a famous "proof" that $\pi=4$, which goes as follows: Start with a square with side-lengths $1$ and inscribe within it a circle with radius $1/2$.
Next, iteratively "fold" the corners of the square in on itself, maintaining the perimeter of $4$, but "approaching" the circle closer and closer (see e.g. this image).
Repeat ad infinitum and we see that the circumference of the inscribed circle, $\pi$, must be $4$.
Clearly, the proof is wrong.
Usually the explanation is that the approaching of the circle is not smooth enough, and so the circumference doesn't follow along.
However, I think the "proof" can be made rigorous using the Hausdorff measure.
Indeed, let $(A_n)_{n\in\mathbb{N}}$ be the sequence of approximating shapes, so $A_1$ is the square with side length $1$, $A_2$ has the corners folded in once, etc.
Let also $C$ denote the circle inscribed within them and $\mu$ the $1$-dimensional Hausdorff measure.
Then clearly, $C$ and $A_n$ are measurable for all $n\in\mathbb{N}$ since these are Borel sets, and since $\mu$ formalizes the circumference of sets in $\mathbb{R}^2$, we have $\mu(C)=\pi$ and $\mu(A_n)=4$ for all $n\in\mathbb{N}$.
Finally, $\bigcap_{n\in\mathbb{N}}A_n=C$ and $A_1\supseteq A_2\supseteq A_3\supseteq\ldots$, so by continuity of measures, we have
$$
\pi
=\mu(C)
=\mu\Big(\bigcap_{n\in\mathbb{N}}A_n\Big)
=\lim_{n\to\infty}\mu(A_n)
=4.
$$
Clearly, something is wrong, but I don't know which of the above steps is erroneous.
Any help is greatly appreciated!
Edit: I realized that while it is true that $\bigcap_{n\in\mathbb{N}}A_n=C$ and $A_1\supseteq A_2\supseteq A_3\supseteq\ldots$, the same is obviously not true of the boundaries of these sets, and since the Hausdorff measure denotes circumferences only on the boundaries of sets, this is where the mistake lies.
Best Answer
There are a couple things wrong with the proof. If $\mu$ is the $1$-dimensional Hausdorff measure and $A_n$ are the solid regions in the plane inside the folded squares, then $\mu(A_n) = \infty$ for every $n$, and $\mu(C) = \infty$ so while it's true that $\mu(\bigcap A_n) = \mu(C)$, it isn't informative.
If $\mu$ is the $1$-dimensional Hausdorff meausre, and the $A_n$ are the boundaries of the solid regions, and $C$ is the boundary of the disk, then it is no longer true that $A_1\supset A_2\supset\dots$ and it certainly isn't true that $\bigcap A_n = C$, so the result $\pi = 4$ of the final calculation using continuity of measures isn't deduced properly.