Measures that coincide on every $(a,b)$ are equal From Bass, Real Analysis for Graduate Students:

Suppose $X$ is the set of real numbers, $\mathcal B$ 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 $A\in \mathcal B$.

I'm thinking the $\pi-\lambda$ theorem comes in handy in this situation. I'm therefore inclined to let $\mathcal C = \{(a,b), a<b\}$ and $\mathcal D = \{A\in \mathcal B, m(A)=n(A)\}$.
$C$ is obviously a $\pi$-system, and I expect $\mathcal D$ to be a Dynkin system (also referred to as $\lambda$-system).
It's easy to prove that $\mathbb R\in \mathcal D$ and $\mathcal D$ is closed under increasing union. Nonetheless, I'm running into some trouble when attempting to prove that $\mathcal D$ is stable under complement...
Indeed, given $A,B$ in $\mathcal D$ such that $A\subset B$ , the following holds:
$m(B) = m(B\setminus A) + m(A)$, hence $n(B) = m(B\setminus A)+n(A)$.
But similarly, $n(B) = n(B\setminus A) + n(A)$.
Hence $m(B\setminus A)+n(A) = n(B\setminus A) + n(A)$
The problem is, $1+\infty = 2+\infty $, yet $1\neq 2$. I cannot derive $m(B\setminus A) = n(B\setminus A)$ unless $n(A)<\infty$, which may not be true...
Note that I haven't used finitess of $n$ and $m$ on $(a,b)$ yet ...
Any help is much appreciated.
 A: Following BigbearZzz's suggestion, let $\mathcal D_k = \{A\in \mathcal B, m(A\cap E_k) =  n(A\cap E_k)\}$.
I'm going to prove that $\mathcal D_k$ is a Dynkin system such that $\mathcal C\subset \mathcal D_k$.


*

*$m(\mathbb R\cap E_k) = m(E_k)=m((-k,k))=n((-k,k))=n(\mathbb R\cap E_k)$ hence $\mathbb R\in \mathcal D_k$

*Let $A,B\in \mathcal D_k$ such that $A\subset B$. Note that  $E_k\cap (B\setminus A) = (E_k\cap B)\setminus (E_k\cap A)$ and both $(E_k\cap B)$ and $(E_k\cap A)$ have finite measure(for both $m$ and $n$). Therefore, $$\begin{align}m(E_k\cap (B\setminus A)) &= m(E_k\cap B)-m(E_k\cap A)\\&=n(E_k\cap B)-n(E_k\cap A)\\&=n(E_k\cap (B\setminus A))\end{align}$$


Hence $B\setminus A\in \mathcal D_k$


*

*If $(A_i)$ is an increasing sequence of elements of $\mathcal D_k$, $$\begin{align}m(E_k\cap (\cup_i A_i)) &= m(\cup_i(E_k\cap A_i))\\&= \lim_im(E_k\cap A_i)\\ &=\lim_i n(E_k\cap A_i)\\ &=n(E_k\cap (\cup_i A_i))  \end{align}$$


Hence $\cup_i A_i \in \mathcal D_k$


*

*$\mathcal C \subset \mathcal D_k$ because $\mathcal C$ is closed under finite intersection.


$\pi-\lambda$ theorem yields $\sigma(\mathcal C)\subset \mathcal D_k$ that is to say $\mathcal B\subset \mathcal D_k$. This means that for any $A\in \mathcal B$, $m(A\cap E_k) = n(A\cap E_k)$.
To finish it off, given $A\in \mathcal B$, $$\begin{align} m(A)&=m(\cup_i (A\cap E_i))\\ &= \lim_i m(A\cap E_i) \\ &= \lim_i n(A\cap E_i) \\&= n(A) \end{align}$$
