Show that two measures are the same I have that $u$ and $v$ are two measures on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ and $u=((-\infty,a]) = v((-\infty,a]) < \infty $ for all $a\in \mathbb{R}$ and I want to show that $u=v$.
I know that the set $(-\infty,a]$ is stable under intersection so I think that I could apply the uniqueness of measures theorem but I am not sure how to actually apply it.
Any inputs on this?
 A: This is more or less how the strategy based on $\lambda$-$\pi$ systems works. I leave some details for the OP.

*

*Consider the restriction $u_n(\cdot)=u(\cdot\cap(-\infty,n])$ and $v_n=v(\cdot\cap(-\infty,n])$. These are finite measures. (why?)


*It is not difficult to check that  $\mathcal{A}_n:=\{B\in\mathscr{B}(\mathbb{R}): u_n(B)=v_n(B)\}$ is a $d$-system (i.e., it contains $\mathbb{R}$; whenever $A,B\in\mathcal{A}_n$ and $A\subset B$, then $B\setminus A\in\mathcal{A}_n$; whenever $\{B_n:n\in\mathbb{N}\}\subset\mathcal{A}_n$ and $B_n\subset B_{n+1}$,  then $\bigcup_nB_n\in\mathcal{A}_n$).


*As $\mathcal{A}_n$  contains the multiplicative system $\mathcal{M}=\{(-\infty,a]:a\in\mathbb{R}\}$ (why?), it follows that $\mathcal{A}_n=\sigma(\mathcal{M})$ by Dynkin's theorem.
Since $\sigma(\mathcal{M})=\mathscr{B}(\mathbb{R})$ (why?) we obtain that
$u_n=v_n$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$.


*To conclude, notice that for any $B\in\mathscr{B}(\mathbb{R})$, $B=\bigcup_{n\in\mathbb{N}}B\cap(-\infty,n]$
I hope this helps.
A: Finite unions of half closed intervals $(a,b]$ form an algebra which generates tht Borel sigma algebra. The hypothesis implies that $u=v$ on this algbra. Also the measures are sigma finite. The uniqueness part of Caratheodorý Extension Thorem now gives the conclusion immediately.
