This is an exercise from a real analysis book that is supposed to help you with entrance exams. I am trying to teach myself.

Suppose $X$ is a set of real numbers, and $B$ is the Boresl $\sigma$-algebra. $m$ and $n$ are two measures on $(X,B)$ such that $m((a,b))=n((a,b))< \infty$ whenever $-\infty<a<b<\infty$. I want to show that $m(A)=n(A)$ for $A\in B$

I have an idea of letting $S$ be a family of sets where the condition of the measures is satisfied, that is $m((a,b))=n((a,b))$. I think showing that S is a $\sigma$-algebra should be enough.

I am having trouble with the complement part because of the interval. I am pretty comfortable working with sets, but the intervals are throwing me off a bit.

Thanks for any input!

  • $\begingroup$ Do you really mean "a set of real numbers," or do you mean "the set of real numbers?" If $X$ doesn't have to be the whole real line, then the condition on the measure of intervals is ambiguous. Leaving that aside: for the complement you get pair of closed intervals, each of which you can write as a disjoint union of open intervals except for a countable set-which must be of measure zero. $\endgroup$ – Kevin Carlson Sep 15 '14 at 4:13
  • $\begingroup$ Hint: Check out the Dynkin $\pi-\lambda$ theorem. If you can show that the collection of sets where $m$ and $n$ agree is a Dynkin system, then, since the open intervals form a $\pi$-system, you'll be done. en.wikipedia.org/wiki/Dynkin_system $\endgroup$ – Josh Keneda Sep 15 '14 at 4:57
  • $\begingroup$ @KevinCarlson Points don't have to be null. $\endgroup$ – Josh Keneda Sep 15 '14 at 5:00
  • 1
    $\begingroup$ Oops, duh, thanks. $\endgroup$ – Kevin Carlson Sep 15 '14 at 5:03

Let $S$ be the set of measurable sets $A$ such that $m(A\cap I)=n(A\cap I)$ for all bounded intervals $I = (a,b)$. We want to show that $S$ contains the Borel $\sigma$-algebra.

We show that $S$ is a Dynkin system. This means that it has to satisfy three properties:

1.) $X \in S$

2.) $S$ is closed under complements

3.) $S$ is closed under countable disjoint unions.

Let $I = (a,b)$. Clearly $(1)$ is satisfied, since $m(X\cap I) = m(I) = n(I) = n(X\cap I)$.

Let $A\in S$. Then we have $$m(A^c\cap I) = m(I) - m(A\cap I) = n(I) - n(A\cap I) = n(A^c \cap I).$$ So $A^c \in S$.

Finally, if $A_1, A_2, A_3, ...$ are disjoint measurable sets in $S$, then $$m(\bigcup_1^\infty A_i \cap I) = \sum_1^\infty m(A_i \cap I) = \sum_1^\infty n(A_i \cap I) = n(\bigcup_1^\infty A_i \cap I).$$

So $S$ is a Dynkin system. It contains the open intervals, since the intersection of any two bounded open intervals is again a bounded open interval (possibly degenerate) on which $m$ and $n$ agree. This also implies that open intervals form a $\pi$-system. Thus, by the Dynkin $\pi$-$\lambda$ theorem, $S$ contains the Borel $\sigma$-algebra. Thus, for any set $A$ in the Borel $\sigma$-algebra, we have: $$m(A) = \lim_{j \rightarrow \infty} m(A \cap (-j, j)) = \lim_{j \rightarrow \infty} n(A \cap (-j, j)) = n(A).$$

  • 2
    $\begingroup$ I am impressed with the friendly analysis you gave here, befitting a learning person like me. I posted similar question, the only difference is that while the OP's question comes from advanced book, mine comes from chapter 3 of Richard Bass' introductory book, so I have to solve it without any "big tools" like Dynkin Theorem. I am wondering if you could help me with my problem here math.stackexchange.com/questions/1135884/… , for there're also people wanting to see solution without big tools. Thank you for your time. $\endgroup$ – Amanda.M Feb 7 '15 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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