How to prove a set belongs to Borel sigma-algebra?

[A.Magnus]

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

Suppose $X$ is the set of real numbers, $\mathcal B$ is the Borel $\sigma$-algebra, and $m$ and $n$ are two measures on $(X, \mathcal B)$ such that $m((a, b))=n((a, b))< \infty$ whenever $−\infty<a<b<\infty$. Prove that $m(A)=n(A)$ whenever $A\in \mathcal B$.​

Here is what I am envisioning but I am not so sure: 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 here is my question:

How do I go ahead proving that $(a, b) \in \mathcal B$? Can I just using the classic formula that if $\forall a \in (a, b) \rightarrow a \in \mathcal B$, then $(a, b) \in \mathcal B$? Any other step I need to follow?​

Thanks for your time and effort.

 

[pasmith]

The Borel $\sigma$-algebra on a topological space is by definition the algebra generated by the open subsets of that space. Since $(a,b) \subset \mathbb{R}$ is open it's in the Borel $\sigma$-algebra of $\mathbb{R}$.

 

[Stephen Tashi]

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.

Why would you be done after that step?

 

[A.Magnus]

The Borel $\sigma$-algebra on a topological space is by definition the algebra generated by the open subsets of that space. Since $(a,b) \subset \mathbb{R}$ is open it's in the Borel $\sigma$-algebra of $\mathbb{R}$.

Yes, I also thought along that line of reasoning similar to yours, but it looks like too easy to be true, therefore I am not so sure about it. Thanks again.

 

[A.Magnus]

Why would you be done after that step?

My reasoning is that since $(a, b) \in \mathcal B$ and since $A$ is an arbitrary set of $\mathcal B$, therefore $m((A)) = n((A))$. Let me know if it is flawed. Thanks again.

 

[Dick]

My reasoning is that since $(a, b) \in \mathcal B$ and since $A$ is an arbitrary set of $\mathcal B$, therefore $m((A)) = n((A))$. Let me know if it is flawed. Thanks again.

I don't follow that at all. You are only given that the two measures are equal for sets that are open intervals. Not all of the sets that are elements of $\mathcal B$ are open intervals.

 

[A.Magnus]

I don't follow that at all. You are only given that the two measures are equal for sets that are open intervals. Not all of the sets that are elements of $\mathcal B$ are open intervals.

My reasoning was shaky at best to begin with, for that reason I posted this question here. I did receive some input on solution to this problem, but all of them requiring big-tool theorems such as Dykin's $\pi - \lambda$ theorem, measurable functions, etc., all of them are out of the range for the time being. In fact this question comes only from the 3rd. chapter of Richard F. Bass' online book http://homepages.uconn.edu/~rib02005/rags010213.pdf on entry-level analysis, therefore those big-tools are not yet in the background.

I am totally lost but I am still hopeful I can get a solution. Thanks again.

 

[Dick]

My reasoning was shaky at best to begin with, for that reason I posted this question here. I did receive some input on solution to this problem, but all of them requiring big-tool theorems such as Dykin's $\pi - \lambda$ theorem, measurable functions, etc., all of them are out of the range for the time being. In fact this question comes only from the 3rd. chapter of Richard F. Bass' online book http://homepages.uconn.edu/~rib02005/rags010213.pdf on entry-level analysis, therefore those big-tools are not yet in the background.

I am totally lost but I am still hopeful I can get a solution. Thanks again.

My measure theory is pretty rusty. But I think you can also use the Monotone Class theorem instead of Dynkin's theorem. That's in the 2nd chapter. Hint: let $\mathcal M$ be the set of all sets such that $m(A)=n(A)$. You'll want to prove that's a monotone class. Use countable additivity of the measures.