Give a construction of three random variables $X,Y,Z$ that are each uniform on $(0,1)$ but $X+Y+Z$ is a constant.
Is the following argument correct? We first consider every number in $(0,1)$ in ternary and for the $n$-th digit, randomly assign $0,1,2$ to $X, Y, Z$, such that all $n$-th digits are different. This means each random variable is uniformly distributed since each of it's digits has a 1/3 chance for each $ \{0,1,2 \}$ To rigorize, you can show that $P(X\leq a) = a$ where $a = k/3^n$ and $n,k$ are positive integers.
So this extends to $P(X\leq r) = r, r\in \mathbb{Q}\cap (0,1)$ because we can find a decreasing sequence $a_j = \frac{k_j}{3^{n_j}}$ s.t $a_j\downarrow r$. Do I need to justify this step more? And then from here we extend it to reals in the same way. We use decreasing sequences because the CDF is right continuous. This shows that $X$ is uniform and similarly $Y$ and $Z$ are uniform on $(0,1)$ They add up to a constant since the digit-wise sum is just $0+1+2$ for each place.
I think this construction is a little bit easier: Let $X$ be uniform on (0,1) and define
$$ Y = \begin{cases}X+\frac 12, \quad X<\frac 12, \\ X- \frac 12, \quad X \geq \frac 12, \end{cases} $$ and
$$ Z = \frac 32 - (X+Y). $$ Then clearly, $X+Y+Z = \frac 32$ and it remains to show that $Y$ and $Z$ are uniform on $(0,1)$.
A simple calculation shows that for $y \in [0, \frac 12)$ $$ \Bbb{P}(Y \leq y) = \Bbb{P}\biggl(X \in \Big[\frac 12, \ y+ \frac 12\Bigr]\biggr) =y, $$ and similarly for $y \in [\frac 12 , 1]$ $$ \Bbb{P}(Y \leq y) = \Bbb{P}\Bigl(X \geq \frac 12\Bigr) +\Bbb{P}\Bigl(X \leq y - \frac 12 \Bigr) =y, $$ so $Y$ is also uniform on $[0,1]$.
Finaly note that
$$ Z = \frac 32 - (X+Y) = \begin{cases}-2X+ 1, \quad X<\frac 12, \\ -2X + 2, \quad X \geq \frac 12, \end{cases} $$ and we can similarly one can show that $Z$ has a uniform distribution on $(0, 1)$.