Currently I am taking a course in probability and encountered the following exercise that deals with different types of convergence of random variables.
The question is as follows. Let $(\Omega, \mathscr{F}, P) = ([0, 1), \mathscr{B}_{[0, 1)}, \lambda_{[0, 1)})$ be a measure space and $X_n = \sqrt{n}\mathbb{1}_{(\frac{1}{n}, \frac{2}{n})}$ be a sequence of random variables. Examine whether the sequence converges in probability, almost surely and convergence in $\mathscr{L}^p$ respectively.
What I know are the relevant definitions.
$(X_n)_{n\geq 1}$ converges to $X$ in probability $:\Leftrightarrow$ $\forall \epsilon > 0$: $\lim\limits_{n\to \infty} P(\vert X_n - X \vert \geq \epsilon) = 0$
$(X_n)_{n\geq 1}$ converges to $X$ almost surely $:\Leftrightarrow$ $P(\lim\limits_{n\to\infty} X_n = X) = 1$
$(X_n)_{n\geq 1}$ converges to $X$ in $\mathscr{L}^p$ $:\Leftrightarrow$ $\lim\limits_{n\to\infty} \mathbb{E}[\vert X_n - X\vert^p] = 0 $
I also know that I can prove convergence a.s. or convergence in $\mathscr{L}^p$ then convergence in probability follows.
What I don't know is how to apply all these definitions practically. I was only given the exercise and we did not have any examples in the lecture.
What I tried is the following. First, I have to figure out the limit $X$. As the set of the indicator function, ($\frac{1}{n}, \frac{2}{n}$), encompasses a smaller and smaller interval with $n \to \infty$ and $\lim\limits_{n \to \infty} (\frac{1}{n}, \frac{2}{n}) = \{0\}$ I assumed $X = 0$. If so, then for convergence in probability we would have
$$\lim\limits_{n \to \infty} P(\vert X_n - X\vert \geq \epsilon) = \lim\limits_{n \to \infty} P(\vert X_n \vert \geq \epsilon) = ?$$
I know what the $X_n$ are but I don't see what probability distribution it is assigned?
Similarly, for convergence a.s. I would, according to the definition, need $\lim X_n = \lim\limits_{n\to\infty} \sqrt{n}\mathbb{1}_{(\frac{1}{n}, \frac{2}{n})}$ but this seems to become infinitely large in $\{0\}$ and zero everywhere else. But, as $\{0\}$ is a set of measure zero, I would conclude convergence almost surely as this excludes sets of measure zero. Correct? If so, then I could conclude convergence in probability as well (though, I would like to see how to prove this with the definition).
Finally, for convergence to 0 in $\mathscr{L}^p$ I consider
$$\lim\limits_{n\to \infty} \mathbb{E}[\vert X_n \vert^p] = \int (\sqrt{n}\mathbb{1}_{(\frac{1}{n}, \frac{2}{n})})^p dP = \int\limits_{(\frac{1}{n}, \frac{2}{n})} n^{\frac{1}{2} + p} dP = n^{\frac{1}{2} + p}\cdot\frac{1}{n} = n^{p - \frac{1}{2}}$$
Hence, I would conclude that this converges for $n \to \infty$ for all $p \geq 1$ and therefore there is no convergence in $\mathscr{L}^p$. Is this correct?