# How to prove that if $\mu(A)=\frac{1}{2}$ for all sets $A$ of the Lebesgue measure $\frac{1}{2}$, then $\mu$ is the Lebesgue measure itself?

Assume that $\mu$ is a Borel measure on $[0,1]$ and let $\lvert\, \cdot\, \rvert$ be the Lebesgue measure on $[0,1]$. Suppose that for any Borel set $A\subset[0,1]$ with $\lvert A\rvert=\frac{1}{2}$ we have $\mu(A)=\frac{1}{2}$. Prove that $\mu=\lvert\,\cdot\,\rvert$.

This is my homework and I do not even know how to start. I suppose I should see the connection with the $\pi-\lambda$ method, but I cannot find any $\pi$–system in this problem. Does the collection of all Borel sets of the Lesbesgue measure $\frac{1}{2}$ generate the Borel $\sigma$–algebra? I would be grateful for your help.

## 2 Answers

Hint: Let $A_1$ and $A_2$ be two disjoint Borel sets with $|A_1|=|A_2|=a$ for some $a$ in $[0,\frac12]$. There exists a Borel set $B$ disjoint from $A_1\cup A_2$ with measure $|B|=\frac12-a$. Let $B_i=A_i\cup B$. Then $|B_1|=|B_2|=\frac12$ hence $\mu(B_1)=\mu(B_2)$. But $\mu(B_i)=\mu(A_i)+\mu(B)$ hence $\mu(A_1)=\mu(A_2)$.

In particular, $\mu\left(\left(\frac{k-1}n,\frac{k}n\right]\right)$ does not depend on $1\leqslant k\leqslant n$ (why?), for each $n\geqslant2$. This proves that $\mu\left(\left(\frac{k-1}n,\frac{k}n\right]\right)=$ $\ldots$

Can you take it from here?

• Clearly, $\mu([0,1])=1$. Applying your argument with $A_1$ and $A_2$, I get $\mu((\frac{k-1}{n},\frac{k}{n}])=b$ for all $k$. Also, $\mu([0,1])=\mu(\bigcup\limits_{k=1}^n \left(\frac{k-1}{n},\frac{k}{n}\right])=nb$, so $b=\frac{1}{n}$ and $\mu$, $\lvert\,\cdot\,\rvert$ agree on the class of all open-closed intervals that is $\pi$–system which generates the desired $\sigma$–algebra. Is that all ok? – Kuba Helsztyński Oct 29 '12 at 20:33
• Yep. Well done. – Did Oct 29 '12 at 20:36
• Thank you so much for all your help. – Kuba Helsztyński Oct 29 '12 at 20:38

Hint: It suffices to show that $\mu$ is equal to Lebesgue measure on every open subinterval. If you partition $[0,1]$ into $2^n$ subintervals of equal length, you can regroup them so that you can use your condition to show their measure equals Lebesgue measure. Approximate open intervals with such sets from within.