Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

Hint: Let $A_1$ and $A_2$ be two disjoint Borel sets with $|A_1|=|A_2|=a\leqslant\frac12$. There exists a Borel set $B$ disjoint from $A_1\cup A_2$ with measure $\frac12-a$. 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 $n\geqslant2$. Hence $\mu\left(\left(\frac{k-1}n,\frac{k}n\right]\right)=$ $____$.

Can you take it from here?

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.