• The key idea is that the proof implicitly relies on switching a limit with an integral with a sleight of hand:
• Let $b(t)$ where $t \in [0, 1]$ be a arclength parametrization of the circle's boundary.
• Then the permiter is $P = \int_0^ b(t)$.
• The troll proof depends on creating a sequence of functions $b_i(t)$, where $\int_0^1 b_i(t) = 4$ for all $b_i$, such that $b_i \to b$ pointwise, but not uniformly.
• Intuitively, why do we need uniform convergence here

§ Another example: $x \mapsto x^n$ on $[0, 1]$

• The sequence of functions converge to $f(x) = 0$ for $x < 1$ and $f(1) = 1$.
• Every function in the sequence is continuous, but the limit function is not.
• If we want to be sure that we can riemann integrate the limiting function, it's "good" if it is continuous.
• Thus, we need to stipulate that the limit ought to be continuous, and we can get this by asking for uniform continuity.

§ Think of it in a fibrational sense

• Given a sequence $f_n(z) : \mathbb R \to D^1$, we can think of a product space $R \times D^1$, the cylinder.
• Then, for each $z \in \mathbb R$ (height level set of cylinder), we get scattered points for $f_n(z)$.
• For each $n \in \mathbb N$, we can plot $f_n(\mathbb R)$, which will give us a curve corresponding to the nth function.
• Pointwise convergence says that for each $z \in \mathbb R$ (for each height level set), the points will converge to the limit point $f(z)$.
• Universe convergence says that if we view the threads $f_n(\mathbb R)$ as a whole, we can make the threads through the cylinder get as close to the function $f$ as we want. This is because for a given bound $\epsilon$, we can apply it uniformly at all heights $z$(that's the uniform bit), so we can squeeze all the threads together to read the limit thread $L$!
• This is nice, because by writing it as a produce space, we can visualize what both parameters $n$ and $z$ are doing.
• The $n$ indexes the various threads, and the $z$ indexes moving through the cylinder.
• math.se link

§ Generalizing the pi troll

• Given a fuction $F : [0, 1] \to R$ such that $\int_0^1 F = N$, and an arbitrary value $M$, design a sequence of functions $f_i$ such that (a) $f_i$ converges pointwise to $F$, and $\int_0^1 f_i = M$.
• The key idea of the construction is to make:
• $f_0$ the zero function
• $f_1$ the function that equals $F$ in $[0, 1/2]$, rescaled to make integral equal $M$, and $0$ in $[1/2, 1]$
• $f_2$ is the function that equals $F$ in $[0, 3/4]$, rescaled to make integral equal $M$, and is $0$ in $[3/4, 1]$
• $f_3$ is the function that equals $F$ in $[0, 7/8]$, rescaled to make integral equal $M$, and is $0$ in $[7/8, 1]$.
• And so on.
• Explicitly, we will have $f_1 \equiv \frac{M}{\int_0^{1/2} F} f \cdot I_{[0, 1/2]}$.
• Explicitly, we will have $f_2 \equiv \frac{M}{\int_0^{3/4} F} f \cdot I_{[0, 3/4]}$.
• Explicitly, we will have $f_3 \equiv \frac{M}{\int_0^{7/8} F} f \cdot I_{[0, 7/8]}$.