I am slightly confused with this borderline case regarding $L^p$ convergence.
In some probability books, they clearly state that $p<\infty$ whereas the online sources do not impose this restriction.
We know that $L^p$ convergence implies convergence in probability. The proof uses Chebyshev's inequality as follows:
$P(|X_n - X| \ge \epsilon) \le \frac{E(|X_n - X|^p)}{\epsilon^p} \rightarrow 0$ as $n \rightarrow \infty$ for $\epsilon > 0$. Does this proof work for $p = \infty$? I am curious how one would go about it if $\epsilon < 1$.
Also, we know that $L^p$ convergence implies that $E(X_n^q) \rightarrow E(X^q)$ for all $q \le p$ by the embedding of $L^p$ spaces. Does this mean that if $X_n$ converges to $X$ in $L^{\infty}$ then all moments of $X_n$ converge to that of $X$ ?
It just seems hard to believe from a simulation perspective since $L^{\infty}$ norm can be approximated by maximum of samples of $|X_n-X|$ whereas for $p<\infty$, the $L^p$ norm is approximated by averaging samples of $|X_n-X|^p$.
Thoughts? Help appreciated!
If $X_n\stackrel{L^\infty}\longrightarrow X$, then $X_n$ converges almost uniformly to $X$. To see this, let $$E_n=\{|X_n-X|>\|X_n-X\|_\infty\}$$ and $$E=\bigcap_{n=1}^\infty E_n. $$ Let $\varepsilon>0$ and choose $N$ such that $\mathbb P(E_N)<\varepsilon$. Then $\mathbb P(E)\leqslant \mathbb P(E_N)<\varepsilon$. On $E^c$, we have $|X_n-X|\leqslant\|X_n-X\|_\infty$ a.s. so given $\delta>0$, we may choose $N'$ such that $n\geqslant N'$ implies $$\|X_n-X\|_\infty<\delta,$$ yielding $$|X_n-X|<\delta\; (w.p.\ 1). $$
It follows that $X_n$ converges to $X$ in probability (and in fact, almost everywhere).