Let X be the Cantor set, which we view as the space $2^\mathbb{N}$ (the set of all infinite binary sequences), equipped with the product topology. We can construct a Borel probability measure $\mu$ on this space by defining $\mu(C\_{a\_i})=1/2$, where the $C\_{a\_i}=\{x\in X | x\_i=a\_i\}$ are the open subbase cylinders of the product topology, and extending to a $\sigma$-algebra in the standard fashion.


Now, consider the Hilbert space $L^2(X,\mu)$. We can obtain orthonormal bases for it using the measure-space isomorphism between $(X,\mu)$ and $[0,1]$ (with Lesbesgue measure) via the binary decimal representations of real numbers. However, the ordinary bases (e.g., the trigonometric basis) on $L^2([0,1])$ are quite ugly when viewed on the Cantor set.



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> Is there an orthonormal basis for $L^2(X,\mu)$ with nice properties (continuity? simply expressible functions?) relative
>  to the structure of the Cantor set?
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> 
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Nice orthonormal basis for L^2(Cantor set)
Since the Cantor set with your measure is also the compact group $(\mathbb{Z}/2)^\mathbb{N}$ with Haar measure, a natural orthonormal basis is the (continuous) characters $\alpha:X\to S^1$, namely the finite products of coordinates $c\_n(x)$, $n\in\mathbb{N}$ if you view $\mathbb{Z}/2$ as `{-1,1}`. These form the discrete group $(\mathbb{Z}/2)^{(\mathbb{N})}$. 


If you view $X$ as the Cantor middle third, $c\_n(x)$ corresponds to $a\_n(x)-1$, the $n$-th base $3$ digit of $x$ minus 1 (all digits are 0 or 2 by definition). These correspond to Walsh functions mentioned by Willie Wong when you use the measure isomorphism $X\to I$, which maps $x$ to ${1\over2} \sum\_n a\_n(x) 2^{-n}$.


Another possible model is $\mathbb{Z}\_2$, the compact group of [2-adic integers](http://en.wikipedia.org/wiki/2-adic_integers), and the characters are then identified to power-of-two roots of unity, forming a group isomorphic to $\mathbb{Z}[1/2]/\mathbb{Z}$.  ~~This seems to lead to the same basis, although indexed differently.~~ EDIT: as remarked by Greg Kuperberg in a comment, this can't be true.


