Let $f: X \to \mathbb C$ be integrable in a measure space $(X, \mathfrak M, \mu)$, i.e. $\int |f| \, d\mu < \infty$. Suppose that $|f(x)| \leq 1$ for all $x \in X$. How can one compute the limit $$ \lim_{n\to \infty} \int \left ( \frac{f ^n}{1 + n |f|} \right )\, d\mu \quad ? $$
My attempt:
I want to find a Lebesgue integrable $g$ that dominates the sequence $f_n = \frac{f ^n}{1 + n |f|}$ and, then, I would conclude that
$$ \lim_{n\to \infty} \int \left ( \frac{f ^n}{1 + n |f|} \right ) d\mu = \int \left ( \lim \frac{f ^n}{1 + n |f|} \right ) d\mu = 0, $$ since $|f| < 1$ implies $\lim \frac{f ^n}{1 + n |f|} = 0$.
My problem is in find such function $g$, I can see that $|f_n| < 1$ for each $n$, however the function $g = 1$ does not need to be Lebesgue integrable since $\mu(X)$ maybe $\infty$.
Help?
For all $x\in [0,1]$ and $n\geqslant 2$, $$ \frac{t^n}{1+nt}=\frac{t}{1+nt}t\cdot t^{n-2}\leqslant \frac{t}{1+nt}t\leqslant \frac 1nt $$ hence applying the previous inequality with $t=\left\lvert f(x)\right\rvert$ gives $\left\lvert f_n(x)\right\rvert\leqslant \left\lvert f(x)\right\rvert/n$.