# 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 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.

• Thank you very much for your hint, I have found inductively that μ and m coincide on all intervals of length $1/2^n$, yes borel sets generated by intervals, but could you rigorously explain how you can pass from interval to Borels, are you using regularity? – seriously divergent Jan 6 '14 at 9:23
• The Borel sets form the smallest $\sigma$-algebra containing the open sets. Check that the collection of Borel sets where $\mu$ and $m$ coincide is a $\sigma$-algebra. This simply uses that the measure of an increasing union is its supremum, and (for complements) that both are probability measures. But sure, using regularity of Lebesgue measure is perhaps the easiest approach, as you only have to deal with $F_\sigma$ and $G_\delta$ sets, for which checking that their $\mu$ and $m$ measures coincide is easier than for general Borel sets. – Andrés E. Caicedo Jan 6 '14 at 14:52

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 $\mu = m$.

• Thank you, good proof, $\mu<<m$ is a clever observation! – seriously divergent Jan 6 '14 at 9:12