On Convergence in probability does not imply convergence in mean

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I got the following question Example 7.11 My questions are:

  • To solve the part a, I tried to use Markov's inequality. However, it fails to show that $X_n \rightarrow 0$ since expected value of $X_n$ is $n$. Why is that?
  • Why is it true that $\lim_{n \rightarrow\infty} P(|X_n| \geq \epsilon) = \lim_{n \rightarrow\infty} P(X_n = n^2)$?
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For any fixed positive $\varepsilon$, we have the equality of the events $$\left\{\left|X_n\right|\geqslant \varepsilon\right\}=\left\{X_n= n^2\right\}$$ if $n^2\geqslant \varepsilon$. Indeed, consider such an $n$. If $\left|X_n\right|\geqslant \varepsilon$, then we necessarily have $X_n=n^2$. Conversely, if $X_n=n^2$, then $X_n\geqslant \varepsilon$. Now, to conclude, note that for $n\geqslant \sqrt \varepsilon$, $$\mathbb P\left\{\left|X_n\right|\geqslant \varepsilon\right\}=\mathbb P\left\{X_n= n^2\right\}= \frac 1n.$$