Given $P(t): [0,1]\to [0,1]^2$ a space filling curve, can we calculate $\iint_{[0,1]^2}f(x,y) dxdy$ as $\int_0^1 f(P(t))\,dt$ or something alike?

Question

Given $P(t): [0,1]\to [0,1]^2$ a continuous bijection, can we calculate $\iint_{[0,1]^2}f(x,y)\, dx\,dy$ as $\int_0^1 f(P(t))\,dt$ or something alike?

I'm thinking of the $P(t)$s as peano curves: we know such continuous bijections exist, thus, with a single parameter $t$, we can fill up the entire domain of integration $D\subseteq \Bbb R^2$ and so, I'd think that we should be able to calculate the double integral in the title with a single integral (integrating with respect to $t$).

Is this possible?

E: As discussed in the comments of the only answer, there may be a few annoying technicalities here ($P$ not being a bijection), I'd rather not bother with them, but see if this idea is usable somehow.

I'm mostly interested in the Riemann or R-S integral, but related stuff about the lebesgue integral is also welcome.

Answer 1

Such $P(t)$ does not exist. If it did, it would be a homeomorphism (the inverse would also be continuous), since $[0,1]$ is compact and $[0,1]^2$ is separated. But the two are not homeomorphic; you can disconnect the line segment by removing a point, but you can't do the same to the square.

So the entire thing is philosophy about things that can't exist.

Answer 2

Well first, there is no continuous bijection from $[0,1]$ onto $[0,1]^2$. As of course has already been pointed out several times; your reply that you don't want to worry about that seems very curious - if you simply corrected the question to be something more sensible it would be a good question.

Anyway. Given a continuous surjection $P:[0,1]\to[0,1]^2$, is it true that $$\int_0^1\int_0^1 f(x,y)\,dxdy=\int_0^1 f(P(t))\,dt?$$The answer is of course no for "most" such $P$, but it's yes for some $P$, including one of the standard examples - the answer is yes for the example commonly known as the Hilbert curve.

This says that the Hilbert curve $H$ is measure-preserving, which follows from the fact that $H^{-1}([j2^{-n},(j+1)2^{-n}]\times[k2^{-n},(k+1)2^{-n}])$ is "essentially" (that is, except for a set of measure zero) equal to $[m4^{-n},(m+1)4^{-n}]$.

The Hilbert curve has other nice properties. For example, it's easy to see that a space-filling curve cannot be $Lip_\alpha$ for $\alpha>1/2$, and $H$ is in fact $Lip_{1/2}$. This says to me that $H$ is in some sense a very "efficient" space-filling curve; just as bad as needed to get the job done, no worse.