What does this hint mean and how is it useful to solve the problem? I am doing a problem  on convergence of random variable. There was a hint given, but I am still struggling to understand the hint.
Here is the problem:  

Let $Y_n$ be uniformly distributed on $\{0,1,\ldots,n\}$: $\mathbb{P}(Y_n=k)=\frac{1}{n+1},0\le k\le n$. Prove that $X_n:=Y_n/n\xrightarrow{d}X\sim U[0,1]$ as $n\to\infty$. (where $\xrightarrow{d}$ means convergence in distribution).

Here is the diagram that I tried to draw:


And here is the hint:  

The limiting DF $F(x)=x1(x\in[0,1])+1(x>1)$ (where 1 is the indicator) is continuous, so have to prove: $\forall x\in\mathbb{R},F_{X_n}\to F(x)$ as $n\to\infty$. 

My question are:
1. What does the hint mean by the limiting DF?
2. And what does it mean and how does it come up with $F(x)=x1(x\in[0,1])+1(x>1)$? how is it significant to the proof? how does it relate to the diagram?
Many many thanks in advance!
 A: First, find the distribution function of $Y_n$, i.e. a $F_n$ such that $$
  \mathbb{P}(Y_n \leq x) = F_n(x) \text{.}
$$
You already drew a diagram of those $F_n$, so you'll just have to formalize that somehow. You may want to use that $x \lfloor \frac{x}{n} \rfloor = x$ if and only if $x$ is a multiple of $n$.
From that, find the distribution function $\widetilde{F}_n$ of $\widetilde{Y}_n = \frac{1}{n}$. You'll just have to rescale the $x$-axis to get from $F(n)$ to $\widetilde{F}(n)$, because $$
  \widetilde{F}(x) = \mathbb{P}(\widetilde{Y}_n \leq x) = \mathbb{P}\left(\frac{1}{n}Y_n \leq x\right) = \mathbb{P}(Y_n \leq nx) = F(nx) \text{.}
$$
But you knew that already I guess, because you also drew a nice diagram of $\widetilde{F}$.
The limiting distribution function is the distribution function of the limit of the $Y_n$, i.e. the distribution function of $U[0,1]$. So again, you're looking for an $F$ such that $$
  \mathbb{P}(U[0,1] \leq x) = F(x) \text{.}
$$
Now, $\mathbb{P}(U[0,1] \leq x) = x$ if $x \in [0,1]$. You'll just have to extent that to $\mathbb{R}$ so that the result is still $\geq 0$, monotone and bounded by $1$. There's only one way to do that...
Finally, you have to show, using the $F_n$ and $F$ you found, that $F_n \to F$ pointwise, i.e. that for every $x \in \mathbb{R}$, you have $$
  \lim_{n\to\infty} F_n(x) = F(x) \text{.}
$$
Your diagrams basically already show that too - just imagine what happens if you add more and more of those steps...

