I am reading the mentioned book by Rosenthal, in particular pg 74 where a uniform is constructed from an infinite sum of i.i.d. Bernoulli, i.e., $P(X_i=0)=P(X_i=1)=\frac{1}{2}$ with
$$U=\sum_{k=1}^{\infty}\frac{X_k}{2^k}.$$
I understand where for $0\le j< 2^k$
$$P\left(\frac{j}{2^k}\le U < \frac{j+1}{2^k}\right)=\frac{1}{2^k},$$
this follows since to reach to $\frac{j}{2k}$ must have k independent success/failures in the first $k$ summands. My question lies in the next statement, "By additivity and continuity of probabilities, this implies $P(a\le U<b)=b-a$.
The only way I can make sense of that is
$$P(a\le U<b)=P(0\le U <b)-P(0\le U <a)$$, and consider a $b<1$ such that it can be represented as $\frac{t}{2^k}$ for some $t<2^k$, where $t\in \mathbb{N}_{+}$ then by additivity,
$$P\left(0\le U <b\right)=P\left(\cup_{m=0}^{t-1}\left\{\frac{m}{2^k}\le U <\frac{m+1}{2^k}\right\}\right)=\sum_{m=0}^{t-1}\frac{1}{2^k}=\frac{t}{2^k}=b.$$
For a $b$ that does not take this form, since it can be written as $b=\sum_{i=1}^{\infty}\frac{d_i}{2^i}$,where $d_i$ is either $1$ or $0$ depending on the binary expansion of $b$ .
This is equivalent to $\lim_{n\to \infty}\frac{t_n}{2_n}$, for a sequence $t_n$ where $t_n<2^n$ and is the integer denominator of the partial sum after converting to like terms. Then let $A_n=\cup_{m=0}^{t_n} \left\{\frac{m}{2^n}\le U <\frac{m+1}{2^n}\right\}$ and note $A_n\subset A_{n+1}$
$$P(0\le U < b)=P(\cup_{n=1}^{\infty} A_n)=\lim_{n\to \infty}P(A_n)=\lim_{n\to \infty}\frac{t_n}{2^n}=b,$$
by continuity and additivity of probabilities.
Let $D=\{\frac{j}{2^{n}}\mid n\in\mathbb{N},0\leq j\leq2^{n}\}$. Given $a,b\in D$ with $a<b$, we can always choose a common $n$ such that $a=\frac{i}{2^{n}}$ and $b=\frac{j}{2^{n}}$, for some $0\leq i<j\leq2^{n}$. Then $[a\leq U<b]=[a\leq U<\frac{i+1}{2^{n}}]\cup[\frac{i+1}{2^{n}}\leq U<\frac{i+2}{2^{n}}]\cup\cdots\cup[\frac{j-1}{2^{n}}\leq U<b]$. Note that the union is disjoint, so \begin{eqnarray*} P\left([a\leq U<b]\right) & = & \sum_{k=i}^{j-1}P\left([\frac{k}{2^{n}}\leq U<\frac{k+1}{2^{n}}]\right)\\ & = & \sum_{k=i}^{j-1}\frac{1}{2^{n}}\\ & = & \frac{j-i}{2^{n}}\\ & = & b-a. \end{eqnarray*} Finally, if $a,b\in[0,1]$ with $a<b$, by density of $D$ in $[0,1]$, we can choose sequences $(a_{n})$ and $(b_{n})$ in $D$ such that $a<a_{n+1}<a_{n}<\ldots<a_{0}<b_{0}<\ldots<b_{n}<b_{n+1}<b$ and $a_{n}\rightarrow a$, $b_{n}\rightarrow b$. For each $n$, let $A_{n}=[a_{n}\leq U<b_{n}]$. Observe that $A_{1}\subseteq A_{2}\subseteq\ldots$ and $\cup_{n}A_{n}=[a<U<b]$. By continuity of measure, we have that $P\left([a<U<b]\right)=\lim_{n}P(A_{n})=\lim_{n}\left(b_{n}-a_{n}\right)=b-a$.