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Testing a general cumulative sum

Let $U_{n}$ be a sequence of IID random variables with distribution function $F$. Let $S_{n}=sum_{k=1}^{n}U_{k}$ be the corresponding cumulative sum.
Show that:
(a) $lim_{nrightarrowinfty}frac{S_{n}}{n}=EX_{1}$

Now the answer given for this problem is:
$EX_{1}=int_{0}^{1}t,dt$
for $0leq tleq 1$.
I do not understand why they get this result. I thought $EX_{1}=0$.
EDIT: Using the sequence $U_{1}=0,U_{2}=1,U_{3}=1,U_{4}=2,U_{5}=1,U_{6}=1,ldots$ we get $S_{n}=0+1+2+1+1+2+ldots+n=n^{2}$
What is the idea behind choosing the form as above? I was unable to prove the result using the above form. Can someone explain?

A:

begin{align*}
mathsf{E}X_1&=int_0^1int_0^1Pr[U_1=u_1,U_2=u_2],mathrm{d}u_1,mathrm{d}u_2\
&=int_0^1int_0^{u_1}F(u_1-u_2),
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