Suppose we have a sequence $\{d_n\}$ where all the $d_n$ are either 1 or 0, with equal probability. Let $x=\sum_{n=1}^\infty d_n2^{-n}$. I need to show that $\mathbb{P}(x \in [a,b])=b-a$.
I started by assuming the interval is of the form $[p/2^n,q/2^n]$ with the hope that I can later get arbitrary $a,b$ by using suitable set operations. I then noted that $\mathbb{P}(x \in [p/2^n,q/2^n])=1-\mathbb{P}(x \in [0,p/2^n])-\mathbb{P}(x \in [q/2^n,1])$.
In order to get a feel for the problem, I tried working out $\mathbb{P}(x \in [0,p/2^n])$ for $p/2^n=0.11010011_2$. In this case, an arbitrary binary number lying in the interval $[0,p/2^n]$ can begin in one of 5 different ways:
$0.0_2,0.10_2,0.1100_2,0.1101000_2,0.11010010_2$
I then noted that these numbers have 1,2,4,7 and 8 digits after the decimal respectively, which is the same position that 1s appear in $p/2^n$. This establishes the required result, but I am unsure how to generalise this to arbitrary $p$. I'd be grateful if someone could point me in the right direction.
This is for a measure theory class.
Here is one approach:
I am assuming that the number corresponding to a given $d$ is $x(d) = \sum_{n=1}^\infty d_n { 1\over 2^n}$, that is, the sum starts at $n=1$.
I am also assuming that the $d_n$ are fair and independent. It follows from this that $P[(d_1,...,d_n) \in A] = P[(d_{i_1},...,d_{i_n}) \in A]$ for any $P$ measurable set $A$ and for any set of distinct indices $i_1,...,i_n$. It follows from this that if we let $\theta_k(d) = (d_k,d_{k+1},...)$, then $P[d \in A] = P[\theta_k(d) \in A]$. (By 'it follows' I mean that it can be shown with a little work :-).)
It is straightforward to see that $x(d) \in [0,1]$ for all $d$, and so $P[x(d) \in [0,1]] = 1$.
It is also straightforward to see that $x(d) = 1$ iff $d_n=1$ for all $n$ and so $P[x(d) = 1] = 0$.
Hence $P[x(d) \in [0,1)] = 1$.
Pick some $n$ and let $k \in \{0,...,2^n \}$
Then \begin{eqnarray} P[x(d) \in [{k \over 2^n}, {k+1 \over 2^n}) ] &=& P[2^n x(d) \in [k,k+1) ] \\ &=& P[ 2^{n-1} d_1+\cdots + d_n + \sum_{k > n} d_k {2^n \over 2^k} \in [k,k+1) ] \\ &=& P[2^{n-1} d_1+\cdots + d_n =k \text{ and } \sum_{k > n} d_k {2^n \over 2^k} \in [0,1) ] \\ && \ \ +P[2^{n-1} d_1+\cdots + d_n =k-1 \text{ and } \sum_{k > n} d_k {2^n \over 2^k} =1 ] \\ &=& P[2^{n-1} d_1+\cdots + d_n =k] \\ &=& {1 \over 2^n} \end{eqnarray} Hence $P[x(d) \in [0,{k \over 2^n})] = {k \over 2^n}$, and hence $P[x(d) \in [0, t)] = t$, for $t \in [0,1]$.