Does $X_n \stackrel{L^1}{\rightarrow}1$ for $n\rightarrow \infty ?$ Problem:
Let $X_n : (\Omega, \mathscr{A},P) \rightarrow (\mathbb{R},\mathscr{B}(\mathbb{R}))$ be a sequence of random variables with the given distribution:
$$P(X_n = n) = \frac{1}{n} \text{and} P(X_n = 0)= 1-\frac{1}{n}, n\geq 1$$
Does $X_n \stackrel{L^1}{\rightarrow}1$ for $n\rightarrow \infty ?$
Note:
$\mathscr{A}$ is a sigma algebra and $\mathscr{B}(\mathbb{R})$ is a Borel set on $\mathbb{R}.$
Attempt:
I used the definition for convergence for $L^p$ spaces, which says if
$$\displaystyle\lim_{n \rightarrow \infty} E\{|X_n - X|^p\}=0$$
then $X_n \stackrel{L^p}{\rightarrow}X$.
Initially, i want to calculate $E\{X_n-1\}:$
$$E\{X_n-1\}=(n-1)\frac{1}{n}+(0-1)(1-\frac{1}{n})=1-\frac{1}{n}-1+\frac{1}{n}=0$$
Using the defintiion:
$$\displaystyle\lim_{n \rightarrow \infty} E\{|X_n-1|\}\stackrel{X_n\geq 0}{=}\displaystyle\lim_{n \rightarrow \infty} E\{X_n-1\}=\displaystyle\lim_{n \rightarrow \infty} 0=0$$
I can therefor conclude that $X_n \stackrel{L^1}{\rightarrow}1.$
Is this correct?
 A: To complement the comment of @ChristopheLeuridan and @ZoeAllen.
Concerning your computation, note that (for $n\geq 1$)
$$
\mathbb E[\vert X_n - 1\vert] = (n-1)\cdot \frac{1}{n} + (1-0)\cdot \left(1 - \frac{1}{n}\right) = 1 - \frac{1}{n} + 1 - \frac{1}{n} \to 2.
$$
In particular, it will not converge to 1 in $L^1$. This shows the importance of the absolute value in the expectation!!
Next, what does $(X_n)_n$ converge to? Looking at the definition of $X_n$, you see that it is at $0$ "most of the time" and at $n$ only in $\frac{1}{n}$ of the time. A good conjecture would be to think that $X_n$ converges to $0$. This is true in the "in probability" sense, i.e.
$$
\lim_n \mathbb P (\vert X_n - 0\vert > \epsilon) = 0
$$
for all $\epsilon > 0$. (Clear? I leave the proof to you!)
Keeping in mind that convergence in $L^1$ implies convergence in probability, 0 is the only possible limit. However,
$$
\mathbb E[\vert X_n - 0\vert] = \mathbb E[X_n] = n\cdot \dfrac{1}{n} + 0\cdot \left(1-\dfrac{1}{n}\right) = 1.
$$
Again, $(X_n)_n$ cannot converge to $0$ in the $L^1$-sense.
Conclusion: $(X_n)$ does not converge in $L^1$. This is quite an important example where mass "escapes to infinity". This is one of the reasons why some sequences do not converge in $L^1$. The other main reason is captured by the concept of oscillations. If you want to dig into that, have a look at Young measures!
