[proofplan] Fix an index $n \in \mathbb N$ and use the common pointwise bound to dominate $|f_n|^p$ by the constant function $M^p$ on $E$. Since $E$ has finite [Lebesgue measure](/page/Lebesgue%20Measure), the integral of this constant over $E$ is $M^p\mathcal L^1(E)$, which is finite. Taking $p$-th roots gives both membership in $L^p(E)$ and the stated estimate, and the argument is independent of $n$. [/proofplan] [step:Dominate the $p$-th power by an integrable constant on $E$] Fix $n \in \mathbb N$. Since $f_n: E \to \mathbb R$ is Lebesgue measurable, the map $x \mapsto |f_n(x)|^p$ from $E$ to $[0,\infty)$ is Lebesgue measurable. The common bound gives, for every $x \in E$, \begin{align*} 0 \le |f_n(x)|^p \le M^p \end{align*} By monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) applied to the nonnegative [measurable functions](/page/Measurable%20Functions) $x \mapsto |f_n(x)|^p$ and $x \mapsto M^p$ on $E$, \begin{align*} \int_E |f_n(x)|^p\,d\mathcal L^1(x) \le \int_E M^p\,d\mathcal L^1(x) \end{align*} Since $M^p$ is constant on $E$ and $\mathcal L^1(E)<\infty$, \begin{align*} \int_E M^p\,d\mathcal L^1(x) = M^p\mathcal L^1(E) < \infty \end{align*} [guided] We fix an arbitrary index $n \in \mathbb N$ because the desired estimate must hold for each function in the sequence. The hypothesis gives the same number $M \in [0,\infty)$ for all indices and all points: \begin{align*} |f_n(x)| \le M \end{align*} for every $x \in E$. Since the map $t \mapsto |t|^p$ from $\mathbb R$ to $[0,\infty)$ is continuous and $f_n: E \to \mathbb R$ is Lebesgue measurable, the composition $x \mapsto |f_n(x)|^p$ is Lebesgue measurable on $E$. Also, because $p \ge 1$ and both sides of $|f_n(x)| \le M$ are nonnegative, raising to the $p$-th power preserves the inequality: \begin{align*} 0 \le |f_n(x)|^p \le M^p \end{align*} for every $x \in E$. We now integrate this pointwise inequality over $E$ with respect to one-dimensional Lebesgue measure. Monotonicity of the Lebesgue integral applies because both functions $x \mapsto |f_n(x)|^p$ and $x \mapsto M^p$ are nonnegative and measurable on $E$. Therefore \begin{align*} \int_E |f_n(x)|^p\,d\mathcal L^1(x) \le \int_E M^p\,d\mathcal L^1(x) \end{align*} The right-hand side is the integral of a constant over a finite-measure set. Hence \begin{align*} \int_E M^p\,d\mathcal L^1(x) = M^p\mathcal L^1(E) \end{align*} Since $M < \infty$ and $\mathcal L^1(E)<\infty$, this number is finite: \begin{align*} M^p\mathcal L^1(E) < \infty \end{align*} Thus the $p$-th power of $|f_n|$ has finite Lebesgue integral over $E$. [/guided] [/step] [step:Take $p$-th roots to obtain the $L^p$ estimate] From the previous step, \begin{align*} \int_E |f_n(x)|^p\,d\mathcal L^1(x) \le M^p\mathcal L^1(E) < \infty \end{align*} By the definition of $L^p(E)$, this proves $f_n \in L^p(E)$. By the definition of the $L^p(E)$ norm, \begin{align*} \|f_n\|_{L^p(E)} = \left(\int_E |f_n(x)|^p\,d\mathcal L^1(x)\right)^{1/p} \end{align*} Taking $p$-th roots in the preceding integral inequality, using that the map $t \mapsto t^{1/p}$ is increasing on $[0,\infty)$, gives \begin{align*} \|f_n\|_{L^p(E)} \le \left(M^p\mathcal L^1(E)\right)^{1/p} \end{align*} Since $M \ge 0$, \begin{align*} \left(M^p\mathcal L^1(E)\right)^{1/p} = M\,\mathcal L^1(E)^{1/p} \end{align*} Therefore \begin{align*} \|f_n\|_{L^p(E)} \le M\,\mathcal L^1(E)^{1/p} \end{align*} Because $n \in \mathbb N$ was arbitrary, the conclusion holds for every $n \in \mathbb N$. [/step]