For all Borel $E\subset [0,1]$ s.t $m(E)=1/2$, we have $\mu(E)=1/2$ where $\mu$ is a prob meas, $m$ is Lebesgue meas. Is $\mu=m$ on Borel sets? For all Borel $E\subset [0,1]$ such that $m(E)=1/2$, we have  $\mu(E)=1/2$ where $\mu$ is a probability measure on $[0,1]$ and $m$ is the Lebesgue measure. Does this imply that  measures $\mu$ and $m$ agree on all Borel subset of $[0,1]$?
What if we change the number 1/2 with any other number less than 1?
Any hint or suggestions are welcome
 A: A simple solution is to notice that for all $n$, the assumption implies that $\mu$ and $m$ coincide on all intervals of length $1/2^n$, and therefore on all intervals, but the intervals generate the Borel sets. 
Similarly if $1/2$ is replaced by some other number in $(0,1)$. You easily get a sequence of lengths $\ell_n$ approaching $0$ such that $\mu$ and $m$ agree on all intervals of length $\ell_n$ for all $n$, and therefore on all intervals. 
A: Suppose that for some $\theta \in (0,1)$ that $mE = \theta$ implies $\mu E = \theta$.
The hypothesis implies that $\mu \ll m$, and so $\mu A = \int_A f dm$ for some  $f$. We want to show $f=1$ ae. [$m$].
Let $C = \{ x | f(x) <1 \}$, and suppose $mC >0$. If $mC=1$, then we have an immediate contradiction, so we can suppose $mC<1$.
If $mC \le \theta$, we can choose some $t$ such that $m ( [0,t] \setminus C) = \theta - mC$, and then if we let $C'=C \cup ([0,t] \setminus C$), we have $mC' = \theta$, but $\mu C' < \theta$.
If $\theta < mC$, we can choose some $t$ such that $m([0,t] \cap C) = \theta$. Letting $C'=[0,t] \cap C$ shows that $mC' = \theta$, but $\mu C' < \theta$.
Consequently $mC = 0$ and so $f(x) \ge 1$ ae. [$m$]. Since $\int (f-1)dm = 0$, we conclude that $f = 1$ ae. [$m$], and so $\mu = m$.
