Let $p\geq1, \delta>0$ and $1\leq p_{n}\to p$. Prove that if $E|X|^{p+\delta}<\infty$ then $\|X\|_{p_{n}}\to\|X\|_{p}$.
I know that
$\|X\|_{p_n}$ = $(E|X|^{p_n})^\frac{1}{{p_n}}$
Since $p_{n}\to p$, we can pick any $\epsilon>0$ such that $\exists N\in\mathbb{N},n>N:|pn-p|<\epsilon$. I think I such pick an $\epsilon$ such that $\delta=\epsilon$ but I do not know how to continue from here.
$\int_0^{\infty} |x|^{p_n} f(x) dx = \int_0^1 |x|^{p_n} f(x) dx + \int_1^{\infty} |x|^{p_n} f(x) dx$
Using, $|x|^{p_n} \leq |x|^{p+\delta}$ for $x \in [1,\infty]$ for all $n \geq N$ for some $N$ and $E(|x|^{p+\delta}) < \infty$, by dominated convergenec theorem as mentioned in one of the comments, $\lim_{n \rightarrow \infty} \int_1^{\infty} |x|^{p_n} f(x) dx = \int_1^{\infty} \lim_{n \rightarrow \infty}|x|^{p_n} f(x) dx = \int_1^{\infty} |x|^{p} f(x) dx$
Now since $|x|^{p_n} \leq 1$ for $x \in [0,1]$ for all $n \geq N$ for some $N$ and $\int_0^1 |1|^{p_n} f(x) dx < \infty$, you can apply dominated convergence theorem again to conclude: $\lim_{n \rightarrow \infty} \int_0^{1} |x|^{p_n} f(x) dx = \int_0^{1} \lim_{n \rightarrow \infty}|x|^{p_n} f(x) dx = \int_0^{1} |x|^{p} f(x) dx$
