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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.

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    $\begingroup$ They are not continuous, for example $f$ jumps at $x=1/2$ (it's also not completely clear if you even defined $f$ at $x=1/2$, but of course that's the smaller problem). $\endgroup$ Commented Jan 14, 2020 at 23:52
  • $\begingroup$ Also, you probably want to add the requirement that $f,g$ are not constant a.e. $\endgroup$ Commented Jan 14, 2020 at 23:53
  • $\begingroup$ @ChristianRemling You are right. Constant functions should not be allowed here. $\endgroup$
    – De vinci
    Commented Jan 14, 2020 at 23:57

2 Answers 2

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We can take a Peano curve $\varphi: [0,1]\to [0,1]^2$. More precisely, we want a construction like the one here 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 $f(x)=\varphi_1(x)$, $g(x)=\varphi_2(x)$.

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The binary digits $\epsilon_1(x),\epsilon_2(x),\ldots$ of a number $x$ in $[0,1]$ are independent under Lebesgue measure, so the two random variables defined by $$ f(x) := \sum_{k=1}^\infty {\epsilon_{2k}(x)\over 2^k} $$ and $$ g(x):=\sum_{k=1}^\infty{\epsilon_{2k-1}(x)\over 2^k} $$ are independent, and they are evidently continuous functions of $x$. Both are uniformly distributed.

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    $\begingroup$ The function $f$ is not continuous. Consider the sequence $(x_n)$ where $x _1 = 0.01, x_2= 0.011, x_3 = 0.0111, \ldots$, and so on, in binary. This converges to $0.0111.... = 0.1$. But $f(0.1) = 0$, while $f(x_n)$ converges to $\sum_{k=1}^\infty \frac{1}{2^k} = 1$. $\endgroup$ Commented Jan 28, 2020 at 5:32
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    $\begingroup$ A successful implementation of your strategy would amount to the same as Christian Remling's answer - defining a space-filling curve $[0,1] \rightarrow [0,1]^2$ whose projections are independent random variables. Though it may not be apparent from versions that are drawn using diagrams, Peano's original paper describes his curve using (ternary) digits. $\endgroup$ Commented Jan 28, 2020 at 5:41

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