[proofplan]
We estimate the $L^p$ norm directly from the essential bound. Because the $L^\infty$ norm is an essential supremum, we first use an arbitrary margin $\varepsilon>0$ to obtain a pointwise bound outside a null set. Raising this almost-everywhere bound to the power $p$, integrating over the finite [measure space](/page/Measure%20Space), and then letting $\varepsilon\downarrow 0$ gives the desired inequality.
[/proofplan]
[step:Convert essential boundedness into an almost-everywhere pointwise bound]
Define
\begin{align*}
M:=\|f\|_{L^\infty}.
\end{align*}
Since $f\in L^\infty(E,\mathcal E,\mu)$, we have $M<\infty$. Let $\varepsilon>0$ be arbitrary, and define the measurable exceptional set
\begin{align*}
N_\varepsilon:=\{x\in E: |f(x)|>M+\varepsilon\}.
\end{align*}
By the definition of the essential supremum, $\mu(N_\varepsilon)=0$. Hence, for every $x\in E\setminus N_\varepsilon$,
\begin{align*}
|f(x)|\le M+\varepsilon.
\end{align*}
[guided]
Define
\begin{align*}
M:=\|f\|_{L^\infty}.
\end{align*}
The hypothesis $f\in L^\infty(E,\mathcal E,\mu)$ means precisely that this number $M$ is finite. We do not immediately assert the pointwise bound $|f|\le M$ everywhere, because the $L^\infty$ norm is an essential supremum and therefore ignores sets of $\mu$-measure zero.
To handle this correctly, fix an arbitrary number $\varepsilon>0$ and define
\begin{align*}
N_\varepsilon:=\{x\in E: |f(x)|>M+\varepsilon\}.
\end{align*}
The set $N_\varepsilon$ is measurable because $f:E\to\mathbb R$ is $\mathcal E$-measurable and the map $t\mapsto |t|$ from $\mathbb R$ to $\mathbb R$ is Borel measurable. By the definition of $M$ as the essential supremum of $|f|$, the bound $M+\varepsilon$ is an essential upper bound for $|f|$. Therefore
\begin{align*}
\mu(N_\varepsilon)=0.
\end{align*}
Consequently, for every $x\in E\setminus N_\varepsilon$,
\begin{align*}
|f(x)|\le M+\varepsilon.
\end{align*}
This is the precise form of the essential boundedness hypothesis that can be integrated.
[/guided]
[/step]
[step:Integrate the $p$th power using finite measure]
Since $p\ge 1$, the function $t\mapsto t^p$ is increasing on $[0,\infty)$. Therefore, for every $x\in E\setminus N_\varepsilon$,
\begin{align*}
|f(x)|^p\le (M+\varepsilon)^p.
\end{align*}
Because $N_\varepsilon$ has $\mu$-measure zero, changing a non-negative [measurable function](/page/Measurable%20Function) on $N_\varepsilon$ does not change its integral. Hence monotonicity of the integral gives
\begin{align*}
\int_E |f(x)|^p\,d\mu(x)\le \int_E (M+\varepsilon)^p\,d\mu(x).
\end{align*}
The integrand on the right is constant, so
\begin{align*}
\int_E (M+\varepsilon)^p\,d\mu(x)=(M+\varepsilon)^p\mu(E).
\end{align*}
Since $M+\varepsilon<\infty$ and $\mu(E)<\infty$, this number is finite. Thus
\begin{align*}
\int_E |f(x)|^p\,d\mu(x)<\infty.
\end{align*}
By the definition of $L^p(E,\mathcal E,\mu)$, this proves $f\in L^p(E,\mathcal E,\mu)$.
[/step]
[step:Take roots and remove the auxiliary margin]
By the definition of the $L^p$ norm,
\begin{align*}
\|f\|_{L^p}=\left(\int_E |f(x)|^p\,d\mu(x)\right)^{1/p}.
\end{align*}
Using the preceding estimate and the monotonicity of the function $s\mapsto s^{1/p}$ on $[0,\infty)$, we obtain
\begin{align*}
\|f\|_{L^p}\le \left((M+\varepsilon)^p\mu(E)\right)^{1/p}.
\end{align*}
Since $M+\varepsilon\ge 0$ and $\mu(E)\ge 0$, this becomes
\begin{align*}
\|f\|_{L^p}\le (M+\varepsilon)\mu(E)^{1/p}.
\end{align*}
The number $\varepsilon>0$ was arbitrary. Letting $\varepsilon\downarrow 0$ gives
\begin{align*}
\|f\|_{L^p}\le M\mu(E)^{1/p}.
\end{align*}
Substituting back $M=\|f\|_{L^\infty}$ yields
\begin{align*}
\|f\|_{L^p}\le \mu(E)^{1/p}\|f\|_{L^\infty}.
\end{align*}
This is the desired estimate and completes the proof.
[/step]
