Show that if distribution functions $F_1(x)=F_2(x)$ for $x \in \mathbb{Q}$ then measures $\nu_1=\nu_2$ In my homework I have a problem that says:
Let $v_1$ and $v_2$ be probability measures on $(\mathbb{R},\mathbb{B})$ with distribution functions $F_1$ and $F_2$. Show that if $F_1(x)=F_2(x)$ for $x \in \mathbb{Q}$ then $\nu_1=\nu_2$
$\underline{\text{My approach is the following}}:$
If $F_1$ and $F_2$ are distribution functions then $F_1(x)=\nu((-\infty,x)
))$ and $F_2(x)=\nu((-\infty,x))$.
For any $y \in \mathbb{R} \setminus \mathbb{Q}$ there exists a $x \in \mathbb{Q}$ such that $|x-y|<\epsilon$ for all $\epsilon >0 $ because the rationals are dense in $\mathbb{R}$
Hence we have that any $\epsilon >0$ then
$\forall y \in \mathbb{R}\setminus \mathbb{Q}, \, \exists x \in \mathbb{Q}: |\nu((-\infty,x))-\nu((-\infty,y))| < \epsilon$ where I use open intervals because the distribution functions are only right-continuous.
Is this a plausable argument?
 A: According to the  usual definition of distribution functions $F_1(x)=\nu_1((-\infty,x])$ and this makes $F_1$ right-continuous. Some authors do take $F_1(x)=\nu_1((-\infty,x))$ but then we get a left continuous function. So you have to be clear on this point.
If you take the right continuous version, then, for any $x$, there exists a  sequence of rational numbers $r_n$ decreasing to $x$. Since $F_1(r_n)=F_2(r_n)$ for all $n$ we get $F_1(x)=F_2(x)$ by just taking limits on both sides. But then you have to use some standard measure theoretic facts to conclude that $\nu_1(A)=\nu_2(A)$ for  all Borel sets $A$. 
Finite disjoint unions of intervals of the type $(a,b]$  (including infinite half closed intervals) form an algebra which generates the Borel sigma algebra of $\mathbb R$. So equality on these implies equality for all Borel sets by the Monotone Class Theorem.[Note that $\nu_1((a,b])=F_1(b)-F_1(a)=F_2(b)-F_2(a)=\nu_2((a,b])$ and this extends to finite disjoint unions of half closed intervals]. 
A: You can derive that $F_1(x) = F_2(x)$ for all $x\in \mathbb{R}$ using $\epsilon-\delta$ definition of right continuity, as you trying to do in your attempt.
Nevertheless it is not completely obvious why ${F_1}_{\mid \mathbb{Q}} = {F_2}_{\mid \mathbb{Q}}$ implies $\nu_1=\nu_2$. For this note that class $\mathcal{F}$ of Borel subsets $A\subseteq \mathcal{B}(\mathbb{R})$ such that
$$\nu_1(A) = \nu_2(A)$$
is a $\lambda$-system and moreover, sets $(-\infty,x)\in \mathcal{B}(\mathbb{R})$ for $x\in \mathbb{Q}$ from a $\pi$-system generating $\mathcal{B}(\mathbb{R})$. Since
$$\nu_1((-\infty,x)) = F_1(x) = F_2(x) = \nu_2((-\infty,x))$$
for every $x\in \mathbb{Q}$, we derive that $(-\infty,x)\in \mathcal{F}$. Now by Dynkin's $\pi\lambda$ theorem, we deduce that $\mathcal{F} = \mathcal{B}(\mathbb{R})$.
