Independence of two digits in a uniform$(0,1)$ random variable I having troubles proving this:
Let X be a Uniform$([0,1])$ distributed random variable. And let $X_n$ be the nth digit in the decimal expansion of $X$. 
Prove that if $n \neq m$ then $X_n$ and $X_m$ are independent.
Any help will be appreciated.
 A: Choose $\omega \in [0,1)$ and let $x_n(\omega)$ be the $n$th digit in the decimal expansion of $\omega$.
To avoid ambiguity, we will suppose that we always take the shortest decimal expansion of $\omega$, for example, if $\omega = 0.1$, we use the expansion $0.1\bar{0}$, not $0.0\bar{9}$. (Since the set of ambiguities has measure zero, this doesn't really matter, of course.)
Let $I_k = {1 \over 10^k} \{ 0,...., 10^k-1 \}$, for $k=0,1,...$. Note that $|I_k| = 10^k$.
Suppose $d \in \{0,...,9\}$, then we see that $x_n(\omega) = d$ iff $\omega \in [{d \over 10^n}, { d+1 \over 10^n })+I_{n-1}$.
Hence $p\{\omega | x_n(\omega) = d \} = | I_{n-1} | {1  \over 10^n} = {1 \over 10}$.
If $n_1 < n_2$, we see that 
$x_{n_1}(\omega) = d_1 $ and $x_{n_2}(\omega) = d_2 $ iff
$\omega \in [{d_1 \over 10^{n_1}}, { d_1+1 \over 10^{n_1} }) + I_{n_1-1}$ and 
$\omega \in [{d_2 \over 10^{n_2}}, { d_2+1 \over 10^{n_2} }) + I_{n_2-1}$.
Each of the intervals $[{d_1 \over 10^{n_1}}, { d_1+1 \over 10^{n_1} }) + k_1$ (with $k_1 \in I_{n_1-1}$) contains exactly ${1 \over 10} 10^{n_2-n_1}$ of the intervals $[{d_2 \over 10^{n_2}}, { d_2+1 \over 10^{n_2} }) + k_2$ (with $k_2 \in I_{n_2-1}$),
hence we see that 
$p\{\omega | x_{n_1}(\omega) = d_1, x_{n_2}(\omega) = d_2 \} = |I_{n_1-1}| {1 \over 10} 10^{n_2-n_1} {1 \over 10^{n_2}} = {1 \over 100}$.
The same reasoning applies if $n_1>n_2$, 
hence we have
$p\{\omega | x_{n_1}(\omega) = d_1, x_{n_2}(\omega) = d_2 \} = p\{\omega | x_{n_1}(\omega) = d_1 \} p\{\omega |  x_{n_2}(\omega) = d_2 \}$.
Now suppose $A_1,A_2 \subset \{0,...,9\}$, we have
\begin{eqnarray}
p\{\omega | x_{n_1}(\omega) \in A_1, x_{n_2}(\omega) \in A_2 \} &=& \sum_{k_1 \in A_1} \sum_{k_2 \in A_2} p\{\omega | x_{n_1}(\omega) =k_1, x_{n_2}(\omega) =k_2 \} \\
&=& \sum_{k_1 \in A_1} \sum_{k_2 \in A_2} p\{\omega | x_{n_1}(\omega) =k_1 \} p \{\omega | x_{n_2}(\omega) =k_2 \} \\
&=& \sum_{k_1 \in A_1} p\{\omega | x_{n_1}(\omega) =k_1 \}  \sum_{k_2 \in A_2} p \{\omega | x_{n_2}(\omega) =k_2 \} \\
&=& p\{\omega | x_{n_1}(\omega) \in A_1\} p \{\omega| x_{n_2}(\omega) \in A_2 \}
\end{eqnarray}
It follows that $x_{n_1}, x_{n_2}$ are independent.
A: Suppose WLOG that $n<m$. Let $H$ be an event which does not depend on the value of $X_m$. We will show that $\text{Pr}(X_m=k|H)=\frac{1}{10}$. In particular, take $H$ to be the events $X_n=l$ and the entire set to prove that $X_n$ and $X_m$ are independent.
The events $A_k=(X_m=k)\cap H$ for $k=0,\ldots,9$ are pairwise disjoint (Lebesgue) measurable subsets of $(0,1)$ with union equal to $H$.  We further have that $A_k=A_0+k\cdot 10^{-m}$ (here we use the assumption that $H$ does not depend on $X_m$), so $\text{Pr}(A_k)=\text{Pr}(A_0)$ independent of $k$. Therefore
$$\text{Pr}(X_m=k|H)=\frac{\text{Pr}(A_k)}{\text{Pr}(H)}=\frac{1}{10}$$
