[proofplan] Set $M:=\|f\|_\infty$. Boundedness gives the pointwise estimate $|f(x)| \le M$ for every $x \in E$, and monotonicity of $t \mapsto t^p$ on $[0,\infty)$ turns this into $|f(x)|^p \le M^p$. Integrating this inequality with respect to $\mu$ gives an upper bound for $\|f\|_{L^p(E)}^p$, and taking the increasing $p$-th root gives the desired estimate without dividing by $\mu(E)$. [/proofplan]

[step:Bound the $p$-th power of $|f|$ by the uniform norm]Let $M:=\|f\|_\infty$. Since $f$ is bounded, $M<\infty$. If $E=\varnothing$, then the asserted inequality is immediate because \begin{align*} \int_E |f(x)|^p\,d\mu(x)=0. \end{align*} Assume therefore that $E\neq\varnothing$. By the definition of the [uniform norm](/page/Uniform%20Norm), for every $x \in E$, \begin{align*} |f(x)| \le M. \end{align*} Since $p \ge 1$ and the map $t \mapsto t^p$ is increasing on $[0,\infty)$, it follows that for every $x \in E$, \begin{align*} |f(x)|^p \le M^p. \end{align*}[/step]

[guided]Define the finite constant \begin{align*} M:=\|f\|_\infty. \end{align*} This is finite because $f$ is bounded. If $E=\varnothing$, then integration over the empty set gives \begin{align*} \int_E |f(x)|^p\,d\mu(x)=0, \end{align*} so $\|f\|_{L^p(E)}=0$ and the desired inequality follows from the convention $\|f\|_\infty=0$. Now assume $E\neq\varnothing$. The point of introducing $M$ is that the uniform norm gives a pointwise bound, not merely an integral bound. For each $x \in E$, the definition of the supremum gives \begin{align*} |f(x)| \le M. \end{align*} Both sides are non-negative [real numbers](/page/Real%20Numbers). Since $1 \le p < \infty$, the function $t \mapsto t^p$ is increasing on $[0,\infty)$. Applying this increasing function to the preceding inequality yields, for every $x \in E$, \begin{align*} |f(x)|^p \le M^p. \end{align*} This is the estimate that can be integrated over $E$.[/guided]

[step:Integrate the pointwise bound and take the $p$-th root] The function $x \mapsto |f(x)|^p$ is $\mathcal{E}$-measurable because $f$ is measurable and $t \mapsto |t|^p$ is continuous. By monotonicity of the integral applied to the non-negative [measurable functions](/page/Measurable%20Functions) $x \mapsto |f(x)|^p$ and $x \mapsto M^p$, \begin{align*} \int_E |f(x)|^p\,d\mu(x) \le \int_E M^p\,d\mu(x). \end{align*} Since $M^p$ is constant on $E$, \begin{align*} \int_E M^p\,d\mu(x)=M^p\mu(E). \end{align*} Therefore \begin{align*} \|f\|_{L^p(E)}^p=\int_E |f(x)|^p\,d\mu(x) \le M^p\mu(E). \end{align*} The right-hand side is finite because $M<\infty$ and $\mu(E)<\infty$, so $f \in L^p(E,\mathcal{E},\mu)$. Taking the increasing $p$-th root on $[0,\infty)$ gives \begin{align*} \|f\|_{L^p(E)} \le M\mu(E)^{1/p}. \end{align*} Substituting $M=\|f\|_\infty$ proves \begin{align*} \|f\|_{L^p(E)} \le \mu(E)^{1/p}\|f\|_\infty. \end{align*} [/step]