If higher order moment is finite, then lower order moment finite?

Question

I'm trying to prove the theorem below.

If $E(|X|^n)<\infty$ for some positive integer $n$, then $E(|X^k|)<\infty$ for every positive integer $k$ such that $k<n$

Here's what I've tried.

ref1) Proof that if higher moment exists then lower moment also exists

ref2) Probability and Statistical Inference, 2nd Edition Robert Bartoszynski, Magdalena Niewiadomska-Bugaj

---

[Step 1] set inequality about $|X|^k$ and $|X|^n$

if $|X|\ge1$, then $|X|^k\le|X|^n$, while if $|X|<1$, then $|X|^k\le1$.

consequently, $|X|^k \le \max(1, |X|^n) \cdots (1)$.

---

[Step 2] prove the right-hand side of (1) has finite expectation

if $E(|X|^n)<\infty$, then $E(|X|^n) = \int^{1}_{-1}|X|^nf(x)dx \;+\; \int_{|X|\ge1}|X|^nf(x)dx$

and it follows that $\int_{|x|\ge1}|x|^nf(x)dx<\infty$ as a difference of two finite quantities.

Thus, $E(|X|^k))\le E\left\{\max\left(1, |X|^n\right)\right\} = \int^{1}_{-1}f(x)dx \;+\; \int_{|X|\ge1}|X|^nf(x)dx < \infty$

---

[Question 1] In [Step 1], Why contains "=" in $|X|^k\le1$?

As I think, "If $|X| < 1$, then $|X|^k<1$" is right.

---

[Question 2] In [Step 2], what are 'two finite quantities' referring to?

$E(|X|^n)$ and $\int^1_{-1}|X|^nf(x)dx$?

Answer 1

[Question 1] : (Since k>0,"=" is just an implication)

[Question 2] : we only need to know $E(|X|^n)$ is some finite value for the proof. while $\int^1_{-1}|X|^nf(x)dx <=1$ can be derived.

below is the Proof: given, $E(|X|^n)$ is finite. some $k$, $0<k<=n$.

if $|X|<1$ then $|X|^k≤1$, So $\int^1_{-1}|X|^kf(x)dx <= \int^1_{-1}1f(x)dx <=1$ (by definition)

if $|X|>=1$ then $|X|^k≤|X|^n$, So $\int_{|X|\ge1}|X|^kf(x)dx <=\int_{|X|\ge1}|X|^nf(x)dx <= E(|X|^n)$

from above 2 cases, we get $E(|X|^k)<= 1 + E(|X|^n)$. hence $E(|X|^k)$ is finite.