Let X be the set of real numbers, Let $B$ be is the Borel $\sigma$-algebra, $m$ and $n$ are two measures on $(X,B)$ such that $m((a,b))=n((a,b))< \infty \forall a,b \in \mathbb{R}$. Prove that $m(A)=n(A)$, whenever $A \in B$.
Now, I know that this quetion has been asked a lot of times, but I am not getting satisfactory solution. One answer to this question suggests to use the "Dynkin's $\pi-\lambda$ theorem". Now what I have done is
$L=\{\ C \in B : m(C)=n(C) \}\ $.
$P=\{\ (a,b): a,b \in \mathbb{R} \}\ $.
$P$ is clearly a $\pi$-system. Also $P\subset L$. Now if I can prove that $L$ is $\lambda$-system. What we will have by"Dynkin's $\pi-\lambda$ theorem" is that $\sigma(P) \subseteq L$. But $\sigma(P)=B$, whence $B\subseteq L$. And, $L\subseteq B$, by definition. Hence $B=L$ and we are done. But what I have failed to show is $L$ is a $\lambda$-system. $L$ is closed under disjoint union is fine. But how do I show that $L$ is closed under complementation.
Thanks in advance!!
Under the extra condition that $m(\mathbb R)=n(\mathbb R)<\infty$ this problem will evidently not arise.
For $k=1,2,\dots$ define measures $n_k,m_k$ by $n_k(A):=n(A\cap(-k,k))$ and $m_k(A):=m(A\cap(-k,k))$ for $A\in B$.
Then applying your method we find $n_k=m_k$ for every $k$.
If $E\in L$ then $n(E^c)=\lim_{k\to\infty}n_k(E^c)=\lim_{k\to\infty}m_k(E^c)=m(E^c)$ so $E^c\in L$.