# Can Convergence in probability in this problem be reinforced to Almost sure convergence

$$X_1,X_2,…,X_n$$ 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:$$\frac{S_n}{n}-\mu_n\overset{p}{\to }0$$

$$\frac{S_n}{n}-\mu_n=(\frac{S_n}{n}-E(X_i))+(E(X_i)-\mu_n)$$
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.
• If $X_n \to X$ almost surely and $Y_n \to Y$ almost surely then $X_n+Y_n \to X+Y$ almost surely. This hardly requires any proof. It follows from definition of almost sure convergence and the fact that union of two sets of measure $0$ has measure $0$. – Kavi Rama Murthy Dec 8 '18 at 12:15