$X_1,X_2,…,X_n$ are independently and identically distributed and $E(X_i)$ exists, $\mu_n=E(X_n I(X_n \le n)),S_n=\sum_{i=1}^n X_i$.
Proof:$$\frac{S_n}{n}-\mu_n\overset{p}{\to }0$$
My answer is:
$$\frac{S_n}{n}-\mu_n=(\frac{S_n}{n}-E(X_i))+(E(X_i)-\mu_n)$$
According to the law of large numbers, $\frac{S_n}{n}-E(X_i)\overset{p}{\to }0$, and it is easy to proof that $E(X_i)-\mu_n \to 0$, so the proposition is proved.
Because its form is also very close to the strong law of large numbers, I wonder if $\frac{S_n}{n}-\mu_n\overset{a.s.}{\to }0$, too.
Assumptions: The following proposition has been proven that $$X_n\overset{p}{\to }X,Y_n\overset{p}{\to }Y \Rightarrow X_n+Y_n\overset{p}{\to }X+Y$$ It is also established for subtraction, multiplication, and division. However, I don't know if it is established when $X_n,Y_n$ converge almost surely.