Does there exists two non-constant continuous functions $f,g:[0,1]\rightarrow \mathbb{R}$ so that they are in independent (in probability sense) when viewed as random variables over the measure space $([0,1],\mathrm{Lebesgue},\mathcal{B}\_{[0,1]})$, where $\mathcal{B}\_{[0,1]}$ is the Borel sigma field of $[0,1]$? 


I was trying the following thing. Let $U(x)=x$ be be a function on $[0,1]$. This follows uniform distribution. Now I am taking binary exapnsion of $U$. It can be proved that 
$$U = \sum\_{k\geq 1}B\_k/2^k,$$
where $B\_k$'s are i.i.d. $\mathrm{Bernoulli}(1/2)$. In particular $$B\_k(x)=\mathbb{1}(k^{\text{th}}\;\text{binary digit of $x$ after decimal point is 1}).$$ Here $\mathbb{1}(\cdot)$ is the indicator function. Now I am defining
$$f=\sum\_{k\;\text{is odd}}B\_k/2^k,\quad g=\sum\_{k\;\text{is even}}B\_k/2^k.$$ 
Thus $f$ and $g$ are independent but I can not prove they are continuous, though they are almost surely continuous w.r.t lebesgue measure. Any help will be appreciated. Please provide any other example if available. Thank you.



Two independent function when considered as random variable over $([0,1],\mathrm{Lebesgue},\mathcal{B}_{[0,1]})$
We can take a Peano curve $\varphi: [0,1]\to [0,1]^2$. More precisely, we want a construction like the one [here](https://en.wikipedia.org/wiki/Space-filling_curve#Outline_of_the_construction_of_a_space-filling_curve) that spends its fair share of the time in each dyadic square when traced out at unit speed. This will make sure that the image measure of Lebesgue measure on $[0,1]$ under $\varphi$ is (two-dimensional) Lebesgue measure again.


Of course, on the square we have random variables as desired (the coordinates), and we can now pull back to $[0,1]$ and take $f(x)=\varphi\_1(x)$, $g(x)=\varphi\_2(x)$.


