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.