Integral of the coordinate functions of the Peano space-filling curve Let $x:=(x_{1},\ldots,x_{n}):[0,1]\longrightarrow [0,1]^{n}$ be the $n$-dimensional Peano space-filling curve. I want to compute
$$
\int_{0}^{1}x_{i}(t)dt
$$
for each $i=1,\ldots,n$. By using the computer, seems that the above integral is equal to $1/2$. I have tried the following.
According to Sagan's book "Space-Filling Curves", if we define for each $k\geq 1$ and given $1\leq i\leq n$ the function $p_{k,i}:[0,1]\longrightarrow [0,1]$ as
$$
p_{k,i}(t):=3^{kn}\big[ \big(t-\frac{j}{3^{kn}} \big) x_{i}\big( \frac{j+1}{3^{kn}}\big)- \big(t-\frac{j+1}{3^{kn}} \big) x_{i}\big( \frac{j}{3^{kn}}    \big)  \big], 
$$
for $t\in [   \frac{j}{3^{kn}},\frac{j+1}{3^{kn}}  ]$, $j=0,1,\ldots,3^{kn}-1$, then $ p_{k,i}\rightarrow x_{i}$ (the convergence is uniform).
So, we have
$$
\int_{0}^{1}x_{i}(t)dt=\lim_{k}3^{kn}\sum_{j=0}^{3^{kn}-1}\int_{\frac{j}{3^{kn}}}^{ \frac{j+1}{3^{kn}} }p_{k,i}(t)dt=\lim_{k}\big[ \frac{1}{2 \cdot 3^{kn} } + \frac{1}{   3^{kn} }\sum_{j=1}^{3^{kn}  }  x_{i}\big(  \frac{ j-1 }{ 3^{kn} } \big) \big]=\lim_{k}  \frac{1}{   3^{kn} } \sum_{j=1}^{3^{kn}  }  x_{i}\big(  \frac{ j-1 }{ 3^{kn} } \big) .
$$
I do not know how to continue...
Many thanks in advance for your comments.
 A: This is a very interesting question, and an interesting article linked to it, though it's rather technical. It was a good idea the authors summed up the relevant ideas in an introduction. Those coordinate functions are self affine, and (as is mentioned immediately before Theorem 3) $x_i(0)=0,\quad x_i(1)=1$, so I suspect one could prove $x_{i}(t)=1-x_{i}(1-t)$. But that would be technical and tedious, and I'll admit I'm as lazy as everybody else. But Theorem 3 does easily turn the trick: "Each coordinate function of the $n$-dimensional Peano curve $\alpha$ is uniformly distributed." Before that, they wrote "We recall that a function $f$ defined in an interval $I \subset R$ is said to be uniformly
distributed (with respect to the Lebesgue measure $\mu$) if, for any measurable set
$A \subset R,\quad f^{−1}(A)$ is measurable and $\mu(f^{−1}(A)) = \mu(A)$."
With $f=x_i$, this means $$\int^1_0x_i(t)\,dt=\int^1_0x_i(t)\,d\mu(t)=\int^1_0t\,d\mu(t)=\int^1_0t\,dt=\frac12,$$ since Lebesgue integral and Riemann integral coincide, if both exist.
