Limit of quotient of two Lebesgue-integrals Let $(\Omega,\mathcal{A},\mu)$ be a measure space with $0<\mu(\Omega)<+\infty$ and $f \in \mathcal{L}^\infty(\Omega,\mathcal{A},\mu,\mathbb{C})$ with $\|f\|_\infty>0$, where $\|f\|_\infty$ denotes the essential supremum of $f$.
Show that
$$ \lim_{n\to\infty} \frac{\int_\Omega |f|^{n+1}\,\mathrm{d}\mu}{\int_\Omega |f|^n\,\mathrm{d}\mu} = \|f\|_\infty.\\$$
Hint: Factor out a suitable power of $\|f\|_\infty$ in the numerator and in the denominator.

My proof is nearly complete but I don't know what to do in the case of $\mu([|f|=\|f\|_\infty])=0$. 
Just for notation purposes, for $y\in\mathbb{R}$, the set $[|f|=y]$ is defined to be $\{x\in\Omega:\,|f(x)|=y\}$. Analogously, we define $[|f|\le y],\;[|f|<y],\;[|f|\ge y]$ and $[|f|>y]$.
Here would be my (incomplete) proof:
We start by using the given hint and write
$$ \lim_{n\to\infty} \frac{\int_\Omega |f|^{n+1}\,\mathrm{d}\mu}{\int_\Omega |f|^n\,\mathrm{d}\mu} = \lim_{n\to\infty} \frac{\|f\|_\infty \int_\Omega {\left(\frac{|f|}{\|f\|_\infty}\right)}^{n+1}\,\mathrm{d}\mu}{\int_\Omega {\left(\frac{|f|}{\|f\|_\infty}\right)}^n\,\mathrm{d}\mu} = \|f\|_\infty \lim_{n\to\infty} \frac{\int_\Omega {\left(\frac{|f|}{\|f\|_\infty}\right)}^{n+1}\,\mathrm{d}\mu}{\int_\Omega {\left(\frac{|f|}{\|f\|_\infty}\right)}^n\,\mathrm{d}\mu} =: \star\\$$
Now, we want to use the dominated convergence theorem.
For this, we define for $n\in\mathbb{N}$ the function $f_n:\Omega\to [-\infty,+\infty]$ by
$$ f_n(x) = \left(\frac{|f(x)|}{\|f\|_\infty}\right)^n.$$
We know that $[|f|>\|f\|_\infty]$ is a $\mu$-null set.
For $x\in\Omega\setminus[|f|>\|f\|_\infty]$ we have
$$ \lim_{n\to\infty} f_n(x) = \lim_{n\to\infty} \left(\frac{|f(x)|}{\|f\|_\infty}\right)^n =
\begin{cases}
   0 &\mbox{if}\;\; |f(x)|<\|f\|_\infty, \\
   1 &\mbox{if}\;\; |f(x)|=\|f\|_\infty.
\end{cases}
$$
Thus, $f_n \xrightarrow{n\to\infty} \mathbf{1}_{[|f|=\|f\|_\infty]}$ pointwise $\mu$-almost everywhere.
Since $f$ is measurable, we have for all $n\in\mathbb{N}$ that $f_n$ is measurable, too. Indicator functions are measurable and therefore $\mathbf{1}_{[|f|=\|f\|_\infty]}$ is measurable.
For $n\in\mathbb{N}$ and $x\in\Omega\setminus[|f|>\|f\|_\infty]$, we see that
$$ |f_n(x)| = f_n(x) = \left(\frac{|f(x)|}{\|f\|_\infty}\right)^n \le \frac{|f(x)|}{\|f\|_\infty} \le 1 = |\mathbf{1}_\Omega(x)|.$$
And since $\mu(\Omega)<+\infty$, we get $\int_\Omega |\mathbf{1}_\Omega|\,\mathrm{d}\mu = \mu(\Omega) < +\infty$ and hence, $\mathbf{1}_\Omega$ is an integrable function which dominates all $f_n$ on the complement of a $\mu$-null set.
One can simply show that $\lim_{n\to\infty} f_{n+1}(x) = \lim_{n\to\infty} f_n(x)\quad\forall x\in\Omega\setminus[|f|>\|f\|_\infty]$.
Now, we are ready to use the dominated convergence theorem (applied on the sequences $(f_{n+1})_{n\in\mathbb{N}}$ and $(f_n)_{n\in\mathbb{N}}$) and we get
$$ \lim_{n\to\infty} \int_\Omega f_{n+1}\,\mathrm{d}\mu = \lim_{n\to\infty} \int_\Omega f_n\,\mathrm{d}\mu = \int_\Omega \mathbf{1}_{[|f|=\|f\|_\infty]}\,\mathrm{d}\mu = \mu([|f|=\|f\|_\infty]) \le \mu(\Omega) < +\infty.\\
$$
Thus, the sequences $(\int_\Omega f_{n+1}\,\mathrm{d}\mu)_{n\in\mathbb{N}}$ and $(\int_\Omega f_n\,\mathrm{d}\mu)_{n\in\mathbb{N}}$ converge both to $\mu([|f|=\|f\|_\infty])\in\mathbb{R}$, so we are allowed to use the quotient rule for limits and therefore
$$
   \star = \|f\|_\infty \lim_{n\to\infty} \frac{\int_\Omega f_{n+1}\,\mathrm{d}\mu}{\int_\Omega f_n\,\mathrm{d}\mu} = \|f\|_\infty \frac{\lim_{n\to\infty}\int_\Omega f_{n+1}\,\mathrm{d}\mu}{\lim_{n\to\infty}\int_\Omega f_n\,\mathrm{d}\mu} = \|f\|_\infty \frac{\mu([|f|=\|f\|_\infty])}{\mu([|f|=\|f\|_\infty])} = \|f\|_\infty.\\
$$

My question would be: What if $\mu([|f|=\|f\|_\infty])=0$? Then both sequences would converge to $0$, so how would I solve this problem?

Thank you in advance! :)
 A: For an alternative approach, I will refer here to a proof scheme due to R. Schilling. Consider that
$$\|f\|_n:=\bigg(\int |f|^nd\mu\bigg)^{1/n}\geq (\mu(\{\omega:|f(\omega)|>\|f\|_\infty-\varepsilon\}))^{1/n}(\|f\|_\infty-\varepsilon)\stackrel{n\uparrow\infty,\varepsilon \downarrow 0}{\to}\|f\|_\infty$$
Therefore $\liminf_{n \to \infty}\|f\|_n\geq \|f\|_\infty$. Now we show that
$$\mu(\Omega)^{-1/n}\|f\|_n\leq \frac{\int |f|^{n+1}d\mu}{\int|f|^nd\mu}\leq \|f\|_\infty$$
The upper bound is given by
$$\frac{\int |f|^{n+1}d\mu}{\int|f|^nd\mu}\stackrel{\textrm{Hoelder's}}{\leq}\frac{\|f\|_\infty\int|f|^nd\mu}{\int|f|^nd\mu}=\|f\|_\infty$$
The lower bound is given by
$$\mu(\Omega)^{-1/n}\|f\|_n\bigg(\int |f|^n\frac{d\mu}{\mu(\Omega)}\bigg)=\bigg(\int |f|^{n}\frac{d\mu}{\mu(\Omega)}\bigg)^{1/n+1}\stackrel{\textrm{Jensen's}}\leq \int |f|^{n+1}\frac{d\mu}{\mu(\Omega)} $$
where $\mu/\mu(\Omega)$ is a probability measure. Both the $\liminf$ and the $\limsup$ of the original fraction are then bounded between $\|f\|_\infty$ and the claim follows.
