If $\lim_{n \to \infty} \int f^{2n} \, d\mu$ exists, then $|f| \leq 1$ $\mu$-almost everywhere. Let $(X,A,\mu)$ be a finite measure space and $f\in M(A)$ (so $f$ is measurable) en $f^n \in L^1(\mu)$ (Lebesgue integrable) for all $n\geq 1$.
Show that if $\lim_{n \to \infty} \int f^n \, d \mu$ exists and is finite than $|f(x)|\leq$ 1 $\mu$ a.e  
 A: Since
$$\lim_{n \to \infty} \int f^n \, d\mu$$
exists and is finite, we also know that
$$\lim_{n \to \infty} \int f^{2n} \, d\mu = \lim_{n \to \infty} \int |f|^{2n} \, d\mu$$
exists and is finite. If we set
$$A_k := \left\{|f|> 1+\frac{1}{k} \right\},$$
then
$$\limsup_{n \to \infty} \int_{A_k} |f|^{2n} \, d\mu \leq \limsup_{n \to \infty} \int |f|^{2n} < \infty. \tag{1}$$
On the other hand,
$$\int_{A_k} |f|^{2n} \, d\mu \geq \left(1+ \frac{1}{k} \right)^{2n} \mu(A_k).$$
Since $\left(1+ \frac{1}{k} \right)^{2n} \to \infty$ as $n \to \infty$, it follows from $(1)$ that $\mu(A_k)=0$. Consequently,
$$\mu(f > 1) = \mu \left( \bigcup_{k \in \mathbb{N}} A_k \right)=0.$$
Remark: Note that the proof holds true for any measure space (i.e. not necessarily finite).
A: Here is a quicker solution. Simply note
$$
\int \liminf_n |f|^{2n} \, d\mu \leq \liminf \int |f|^{2n} \, d\mu < \infty
$$
by Fatou's Lemma.
But if $M := \{x \mid |f(x)| > 1\}$, then
$$
\liminf_n |f(x)|^{2n} = \infty
$$
holds for all $x \in M$.
Since an integrable function can only assume the value $\infty$ on a set of measure zero, we get $\mu(M) = 0$.
Alternatively, $\infty \cdot \chi_M \leq \liminf_n |f|^{2n}$, where $\chi_M$ is the  indicator function of $M$. Thus
$$
\infty \cdot \mu(M) = \int \infty \cdot \chi_M \, d\mu \leq \int \liminf_n |f|^{2n} \, d\mu < \infty,
$$
which also yields $\mu(M) =0$.
EDIT: Note that we did not use the assumption that $X$ is a finite measure space.
