Continuity and right continuity of two functions I'm trying the following two problems:

Let $(\mu_n)_n$ a sequence of measures on $\mathbb{R}$ such that $\sup_{n\in\mathbb{N}}(\mathbb{R})<\infty$. Prove that exists a right continuous function $F$ and a subsequence such that $\lim_{k\to\infty}\mu_{n_k}((-\infty,x])=F(x)$ for every continuity point $x$ of $F$.

I don't know which function could be appropiate. 
First, I tried $F(x)=\limsup_{n\to\infty}\mu_{n}((-\infty,x])$, but I cannot find the subsequence (This functon is right continuous).
Now, I'm trying this: 
Let $\{q_j\}_{j\geq 1}$ an enumeration of rationals. Consider the sequence $\{\mu_n(-\infty,q_1]\}_{n\geq 1} \subset [0, \sup_{n\in\mathbb{N}}\mu_{n}(\mathbb{R})]$, and a convergent subsequence $\{\mu_{n_k}^{1}(-\infty,q_1])\}_{k\geq 1}$. Now, consider $\{\mu_{n_k}^{1}(-\infty,q_2]\}_{k\geq 1}$ and extract a convergent subsequence $\{\mu_{n_k}^{2}(-\infty,q_2]\}_{k\geq 1}$. We can do this with every $q_j\in\mathbb{Q}$. Then, we choose $\{\mu_{n_k}^{k}\}_{k\geq 1}$, which, by construction, gives the existence of $\lim_{k\to\infty}\mu_{n_k}^{k}(-\infty,q_j]$ for every rational $q_j$. We define $F$ as this limit for $\mathbb{Q}$. Now, for $x\in\mathbb{R}\setminus\mathbb{Q}$, we set $F(x)=\inf\{{F(q_j): q_j>x}\}$. Apparently, this functions is right continuous (I'm improving my demonstration), but, again, I don't know how to prove the subsequence part. 

Let $A, B \subset \mathbb{R}$ Lebesgue-measurable sets of finite measure. For $x\in\mathbb{R}$, let $B+x=\{b+x : b\in B\}$. Prove that $F(x)=\lambda(A\cap (B+x))$ is continuous.

Here, I'm completely lost. 
I really need hints.
 A: The first part is a standard result called Helly's selection principal and almost all books on Probability (including Chung's book) have  a proof. So I will answer the second part. The following is a basic approximation result in measure theory: if $B$ is a Lebesgue measurable set of finite measure and $\epsilon >0$ then there exists a finite disjoint collection $[a_1,b_1),[a_2,b_2),...,[a_n,b_n)$ of half -closed intervals such that  if $C$ is their union then $\lambda (C\Delta B)<\epsilon $. [ Here $ (C\Delta B)=(C\setminus B) \cup (B\setminus C)$]. The idea of the proof of this result is to consider the collection of all Borel sets which can be approximated in this fashion (for every $\epsilon >0$) , show that this class is a sigma  algebra which contains half closed intervals. hence it contains all Borel sets teh result extends easily to Labesgue measurable sets by using just the definition of completion. Once this result is granted you can see that $|F(x)-G(x)|<\epsilon$ for all $x$, where $G(x)=\lambda (A \cap (C+x))$. Now, if $a,b$ and $H(x)=\lambda (A \cap [a,b)+x))$ then $|H(x)-H(y)| \leq \lambda ([a+x,b+x)\Delta [a+y,b+y)) \leq 2|y-x|$. We have proved that the function $F$ can be approximated uniformly by continuous functions. Hence it is continuous.
