 Solution

Convergence in probability means that for every $\epsilon >0$, $\mathbb{P}(|X_n-X|\ge \epsilon )\to 0$. The use of $>$ instead of $\ge$ gives the equivalent definition by replacing $\epsilon$ with a smaller positive number.

For $Y\ge 0$, the pointwise inequality $Y\ge a{1}_{\{Y\ge a\}}$ gives, on taking expected values, Markov's inequalityThis also holds if the expectation is infinite, though the resulting bound is then uninformative. Apply it to $Y=(X-\mu {)}^2$ with threshold ${\epsilon }^2$ to obtain Chebyshev's inequality:Set $S_n=n^{-1}{\sum }_{i=1}^n(Z_i-\mathbb{E}{Z}_i)$. Its mean is zero. Independence eliminates cross covariances, and boundedness gives $\mathrm{Var}(Z_i)\le \mathbb{E}{Z}_i^2\le 1$. HenceThus $\boxed{S_n\to 0\text{ in probability}}$. Identical distributions were never needed.