There is a model of knowledge, essentially due to Robert Aumann, in which knowledge is represented by a partition $\Pi$ of a set of states of the world $\Omega$. If the true state of the world is $\omega$, the agent with partition $\Pi$ only knows that some state in the cell $\pi(\omega)$ (the value of the projection at $\omega$) obtained. An event is simply a subset of $\Omega$. We say that an agent knows that the event $E$ obtains at $\omega$ if $\pi(\omega)\subseteq E$. Now let the state space be $\Omega=\{1,2,\ldots,T\}$, where we interpret $t$ as "there is an exam at $t$". Now there is no partition $\Pi$ such that the following holds:
- The student doesn't know exactly at which date the exam is at any state.
- If there was no exam at $\{1,\ldots,t-1\}$, then the student knows this at $t$.
Proof: Let $t$ be an element in $\Omega$ such that $\pi(t)$ is not a singleton. Such an element must exist by 1. Let $t'$ be the largest element in $\pi(t)$. By assumption $t'>t$ and so by 2., $\{1,\ldots,t'-1\}$ is a union of cells in $\Pi$ that contains $t$. Since $\Pi$ is a partition, $\pi(t)\subseteq\{1,\ldots,t'-1\}$, contradicting $t'\in\pi(t)$.
So at least using the model of knowledge used above, the surprise exam paradox cannot be formulated coherently.
It is a reasonable question. Although usually if asked to find the area of a region or regions bounded by two graphs what is meant by "bounded" is that the regions all lie within the interior of some circle.
This is analogous to a bounded set on the number line being contained in some interval $[a,b]$. It is completely circumscribed.
However it is possible for to graphs to enclose a finite, yet unbounded region.
There are many examples, but one is as follows.
Find the area of the region "bounded" by the graphs of $y=0$ and $y=\dfrac{x}{x^4+1}$
Here is the graph of the region.
This region is not bounded in the sense stated above. It cannot be contained in the interior of a circle. Yet it has a finite area.
\begin{equation}
\int_{-\infty}^\infty\dfrac{|x|}{x^4+1}\,dx=\int_{0}^\infty\dfrac{2x}{x^4+1}\,dx\\
\end{equation}
Make the substitution $u=x^2$, $du=2x\,dx$ and this becomes
\begin{eqnarray}
\int_{0}^\infty\dfrac{1}{u^2+1}\,du&=&\frac{1}{2}\arctan(u){\Large\vert}_{0}^\infty\\
&=&\left(\dfrac{\pi}{2}-0\right)\\
&=&\frac{\pi}{2}
\end{eqnarray}
Therefore it is acceptable to say that, in a sense, an unbounded region is "bounded" by two graphs so long as the area enclosed is finite.
Best Answer
What you've found is a fundamental property (or perhaps it would be better to say lack of property) of arc length in two dimensions — or going up one dimension, surface area in three dimensions: it's not 'continuous' with respect to small changes in shape.
You don't even need a Koch snowflake or similar fractal curve to see this; give me any shape — for instance, a circle — and give me an arbitrarily small box that its boundary passes through, and I can give you a shape that looks exactly the same outside the box but that has arbitrarily large arc length. For an example of this, imagine starting with the curve $y=\sin^2(\pi x)$ between $x=0$ and $x=1$. Then the exact arc length of this curve is hard to compute, but we can put a lower bound on it: the curve goes from $y=0$ to $y=1$, then from $y=1$ to $y=0$ again, so it must have total length at least $2$. $(^*)$
But now consider the curve $y=\sin^2(2\pi x)$ between $x=0$ and $x=1$. This is two copies of the sine curve placed next to each other, and its $y$ values take the 'path' mentioned above twice, so it must have total length at least $4$. And more generally, $y=\sin^2(n\pi x)$ has arc length at least $2n$ between $x=0$ and $x=1$. But the curve is always within the rectangle $0\leq x\leq 1\pi$, $0\leq y\leq 1$. So imagine that you tell me you want to get a continuous curve with minimum arc length $\ell$ that fits into a box of size (side length) $d$. Then I know that if I create a curve with arc length $\ell/d$ within a $1\times 1$ box, I can scale it by a factor of $d$ and get a curve of arc length $\ell$ within the $d\times d$ box. But I also know that I can create a curve with arc length at least $\ell/d$ in the $1\times 1$ box by using the curve $y=\sin^2(2\pi\ell x/d)$ (for $0\leq x\leq1$). So I can just "paste" this curve in wherever you tell me; I've only modified things within the $d\times d$ box, but I've made the arclength arbitrarily large.
As for what this has to do with 'painting' the snowflake, consider the thickness of the paint you're using to try and paint the sides of it. If you want your paint layer to have a finite thickness $\tau$, then we're not necessarily talking about the Koch snowflake itself, but just a curve that lies within a distance $\tau$ of it everywhere; since the paint 'hides' anything smaller than $\tau$, we can't tell the difference between the two anyway. But for any $\tau$ there are curves that stay within distance $\tau$ of the snowflake but have finite length! So we only need a finite amount of paint to do the painting here. And as your paint gets thinner and thinner, the length that you need to cover will increase — but that's all right, because your paint will spread farther and farther. In the limit, you're talking about 'painting' the infinite length of the snowflake itself — but now your paint layer is infinitely thin, so it shouldn't be too surprising that you can make it stretch as far as you need to. (This is very much analagous to taking a one-square-unit 'bucket' of 2d paint and noting that you can paint a $1$ unit $\times 1$ unit square with it, or a $2$ unit $\times 1/2$ unit rectangle, or a $1000$ unit $\times 0.001$ unit rectangle, or...)
Meanwhile, I can't do the same thing with area (or in the 3d case, volume); it's clear to see that any modifications that are made within a $d\times d$ box can only change the area of the figure (one way or the other) by at most $d^2$. $(^*)$ So if you tell me that you want to increase the area bounded by your shape by a thousand square units but only give me a $1/10\times 1/10$ unit box to do it in, I can fairly tell you that that's impossible.
Incidnetally, this same lack of continuity for arc length is at play in another very similar 'paradox': The staircase paradox, or why $\pi\ne4$ . The fundamental explanation is the same: the fact that two curves are arbitrarily close to each other along their entire spans tells you nothing about how close their arc lengths are.
$(^*)$ Note that I haven't given you proofs of these assertions! You should be skeptical here, because you've already seen that behavior doesn't necessarily match what you might expect where these things are concerned. Fortunately, these assertions are true, but proving them would require going pretty far afield.