Strongly continuous measure [reduced] Let us first give our definitions: the word CHARGE means finitely additive measure


strongly continuous finitely additive measure is nonatomic, but not conversely. Here is the example:
  
how cannot we partitioned $X$ into finite sets $\{b_1,b_2,\dots,b_n\}$ in $\mathscr{F}$, each of measure $<\varepsilon=1/2$. If we assume $\varepsilon=1/2^i$ for $i=1,\dots,n$, there is no partition. But I cannot see why for $\varepsilon=1/2$ there is no partition for $X$.


There is no proper subset of $[0,1/4)\cup[3/4,1]$ in your algebra, so some partition cell containing it must be a superset, which has therfore measure at least $1/2$.
