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I am working on this problem on measure theory like this:

Let $X$ be set of $\mathbb R$, and let $\mathcal B$ be its Borel $\sigma$-algebra, and finally let $\mu_1$ and $\mu_2$ be the two measures on $(X,\mathcal B)$ such that $\mu_1((a,b))= \mu_2((a,b)) < \infty$ whenever $−\infty < a < b < \infty$. Show that $\mu_1(A) = \mu_2(A)$ whenever $A \in \mathcal B$.​

Here is what I was at first thinking: Since $a,b \in \mathbb R$ and since $A$ is an arbitrary subset of $\mathcal B$, so if only I can prove that $(a,b) \in \mathcal B$, then I am done. But I was told by a responder to my posting at Physics Forum here that this reasoning is wrong, since not all sets in $\mathcal B$ are open. I am hitting a deadend again.

Therefore I am posting this question here looking for help, thanks for your time and effort.


POST SCRIPT - 1: I should have mentioned this: This problem comes from the 3rd. chapter of an introductory text by Richard F. Bass here, therefore any solution shouldn't involve any advanced theorems such as Dynkin's, etc. Sorry for this belated info, thanks though to all who have taken time to help.


POST SCRIPT - 2: I finally came up with solution without any advanced theorems, adapted from a solution by @JoshKeneda, who used Dynkin's Theorem. I have submitted this work to my professor, he ok'd it except for (5) because it is true only when the $A_i$'s are pairwise disjoint. Feel free to drop me a message if you have ideas to improve (5). Thanks to all and especially to @JoshKeneda.


DISCLOSURE: This question is very similar to an old MSE posting here, which was put on hold due to being incomplete. My posting has all the correction to the first posting. Always conscientious of community rule and guideline, I have tried avoiding duplication by posting this question elsewhere here and here, but I did not receive any meaningful helps $-$ understandably, as those two outside forums are not specialized in math. Thank you for your understanding.

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  • $\begingroup$ I'm very interested in a complete detailed proof of this without using any big tools, just the basics definitions of measure theory. Perhaps Monotone Class Theorem; @A.Magnus, note that if you remove the restriction of $-\infty < a < b < \infty$ it fails, e.g., try counting measure: see here $\endgroup$ – Robert Cardona Feb 6 '15 at 3:19
  • $\begingroup$ @RobertCardona The Monotone Class Theorem is immediately equivalent to Dynkin's theorem. Apply Monotone Class Theorem with indicator functions. $\endgroup$ – nullUser Feb 6 '15 at 3:25
  • $\begingroup$ @nullUser, thanks! I've been hesitant to use it because I didn't think it was what the author intended us to use, and I could never prove the OP's statement from the statement the author gave for the Monotone Class Theorem: $\mathcal A_0$ an algebra, $\mathcal A$ is the smallest $\sigma$-algebra containing $\mathcal A_0$, $\mathcal M$ is the smallest monotone class containing $\mathcal A_0$, then $\mathcal M = \mathcal A$. $\endgroup$ – Robert Cardona Feb 6 '15 at 3:29
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    $\begingroup$ Without the big tools - that is exactly what I have in mind. I forgot to write down that this question comes the first few chapter of introductory-level measure & integration text, so Dynkin's Theorem is probably not in the background. But the Monotone Class Theorem is quite possible, although I do not know how to apply it. :-( Thanks to both of you! $\endgroup$ – Amanda.M Feb 6 '15 at 3:39
  • $\begingroup$ I would be surprised to find any introductory measure theory book that doesn't contain one or both of Dynkin's Theorem or Monotone class theorem in the first few chapters. $\endgroup$ – nullUser Feb 6 '15 at 4:08
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By Dynkin's $\pi$-$\lambda$ Theorem two finite measures of the same mass that agree on a $\pi$-system agree on the $\sigma$-algebra generated by that $\pi$-system.

To prove this using Dynkin, consider $\Lambda:=\{ B \in \mathcal{B} : \mu_1(B \cap (a,b)) = \mu_2(B \cap (a,b))\}$ is a Dynkin system containing the $\pi$-system of open intervals. Thus $\Lambda$ contains the $\sigma$-algebra generated by open intervals (which is $\mathcal{B}$) and hence $\Lambda = \mathcal{B}$. Thus the restriction of $\mu_1$ and the restriction of $\mu_2$ are identical on any finite interval. An arbitrary set $B$ can be written as the increasing union of $B\cap (-n,n)$ for larger and larger $n$, so the result follows.

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    $\begingroup$ I guess first you need to restrict to some finite open intervals and then extends to the real line. Otherwise, you would get trouble in checking $\Lambda$ is indeed a lambda system. $\endgroup$ – Brian Ding Feb 6 '15 at 3:45
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    $\begingroup$ @nullUser : Thank you! Do you mind proving it using the Monotone Class Theorem instead? The text has the theorem in a chapter prior to this problem. Thank you again. $\endgroup$ – Amanda.M Feb 6 '15 at 3:48
  • $\begingroup$ As @BrianDing pointed our, this is not true in general, i.e. consider the counting measure and the measure that assigns infinite measure to all non-empty sets, these agree on all open sets in $\mathbb{R}$ but are not equal. See here for some cases where the result holds. $\endgroup$ – Anguepa Jul 9 '17 at 8:45
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Let $\mathcal{H}$ be the vector space of bounded measurable functions $h$ such that $\int h d\mu_1 = \int h d\mu_2$. (Check this is a vector space, easy). Clearly $1 \in \mathcal{H}$ (though it is not immediate). Suppose that $h_n \geq 0$ and $h_n$ is a sequence in $\mathcal{H}$ such that $h_n$ increases monotonically to the bounded function $h$. Thus $$\int h d\mu_1 = \lim_n \int h_n d\mu_1 = \lim_n\int h_n d\mu_2 = \int h d\mu_2$$ so $h \in \mathcal{H}.$ Since $\mathcal{H}$ contains the indicators of every set in the $\pi$-system $\Pi$ of finite open intervals, the monotone class theorem implies $\mathcal{H}$ contains all bounded $\sigma(\Pi) = \mathcal{B}$ measurable functions, in particular it contains the indicators $1_B$ for all $B \in \mathcal{B}$. Thus $\mu_1(B)=\mu_2(B)$ for all $B \in \mathcal{B}$.

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  • $\begingroup$ The book hasn't covered measurable functions or monotone convergence theorem at the point in which the problem was mentioned. I came across this problem in Chapter 3 of Richard Bass' book, specifically Exercises 3.9 and 3.10. $\endgroup$ – Robert Cardona Feb 6 '15 at 4:09
  • $\begingroup$ With the Lebesgue measure we know that any borel set can be approximated by open sets with arbitrary precision. E.g. math.stackexchange.com/questions/601333/… so we can just use this. Though this argument in my opinion is much more complicated than either of the two I have given. $\endgroup$ – nullUser Feb 6 '15 at 4:14
  • $\begingroup$ @RobertCardona Also in the book you mention, the monotone class theorem is exercise 2.7, and is the subject of the entire section 2.2. $\endgroup$ – nullUser Feb 6 '15 at 4:16
  • $\begingroup$ Yes, Robert Cardona said it correct: It comes from the 3rd. chapter of Richard F. Bass' book. It should be solvable without the "big tools." Thanks to both of you for trying hard. $\endgroup$ – Amanda.M Feb 6 '15 at 13:25
  • $\begingroup$ @nullUser, that's not a version of the monotone class theorem that I'm familiar with. I was surprised that you said Exercise 2.7 was the monotone class theorem (for functions) because when I proved it, I did so without the monotone class theorem for sets, which I would think naturally, would be required in the proof. I posted a question hoping to clarify this. $\endgroup$ – Robert Cardona Feb 9 '15 at 22:36
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(1) Let $I := (a, b)$, let $S := \{A \in \mathcal B \mid m(A \cap I) = n(A \cap I)$}. We need to prove that $S$ is in the Borel $\sigma$-algebra.

(2) Let $A = X \in \mathcal B$, $$\begin{align} m(X \cap I ) &= m(I) \\ n(X \cap I ) &= n(I) \\ \text{therefore, } m(X \cap I ) &= n(X \cap I ) \\ \end{align}$$

(3) Let $A = \emptyset \in \mathcal B$, $$\begin{align} m(\emptyset \cap I) &= m(\emptyset) = 0 \\ n(\emptyset \cap I) &= n(\emptyset) = 0 \\ \text{therefore, } m(\emptyset \cap I) &= n(\emptyset \cap I). \end{align}$$

(4) For $\forall A \in \mathcal B, A^c \in \mathcal B$, $$\begin{align} m(A^c \cap I) &= m((X \setminus A) \cap I)\\ &= m((X \cap I) \setminus (A \cap I))\\ &= m(X \cap I) - m(A \cap I)\\ &= m(I) - m(A \cap I)\\ \text{similarly, } n(A^c \cap I) &= n(I) - n(A \cap I)\\ \text{therefore, } m(A^c \cap I) &= n(A^c \cap I).\\ \end{align}$$

(5) Let $(A_i)_{i \in \mathbb N} \in A$, $$\begin{align} m(\bigcup _{i = 1}^{\infty} (A_i \cap I)) &= \sum_{i = 1}^{\infty} m (A_i \cap I)\\ &= \sum_{i = 1}^{\infty} n (A_i \cap I)\\ &= n (\bigcup _{i = 1}^{\infty} (A_i \cap I))\\ \end{align}$$

(6) Similarly, let $(A_i)_{i \in \mathbb N} \in A$, $$\begin{align} m (\bigcap _{i = 1}^{\infty} (A_i \cap I)) &= m ((\bigcap_{i=1}^{\infty}A_i) \cap I)\\ &= n ((\bigcap_{i=1}^{\infty} A_i) \cap I)\\ &= n (\bigcap _{i = 1}^{\infty} (A_i \cap I))\\ \end{align}$$

(7) Having proven that $m(A \cap I) = n(A \cap I)$ is in Borel $\sigma$-algebra, now we need to prove that $m(A) = n(A)$:

$$\begin{align} \lim _{k \to \infty} (m(A \cap (-k, k)) &= \lim _{k \to \infty} (m (A)) \\ &= m(A) \\ \text{similarly, } \lim_{k \to \infty} (n(A \cap (-k, k)) &= \lim _{k \to \infty} (n (A)) \\ &= n (A)\\ \text{hence, } n(A) &= m(B). \qquad \blacksquare \\ \end{align}$$

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