[proofplan] We split the proof into the cases $q<\infty$ and $q=\infty$. In the finite-$q$ case, we integrate $|f|^p$ as the product $|f|^p\cdot 1$ and apply Hölder's inequality with conjugate exponents $q/p$ and $q/(q-p)$. In the $q=\infty$ case, the essential supremum bound gives $|f|\leq \|f\|_\infty$ almost everywhere, and the finite measure of $X$ turns that pointwise bound into an $L^p$ bound. [/proofplan]

[step:Apply Hölder to $|f|^p$ and the constant function when $q<\infty$]Assume first that $q<\infty$. Define the exponents $r,s\in[1,\infty]$ by \begin{align*} r := \frac{q}{p} \end{align*} and \begin{align*} s := \frac{q}{q-p}. \end{align*} Since $1\leq p<q<\infty$, we have $1<r<\infty$, $1<s<\infty$, and \begin{align*} \frac{1}{r}+\frac{1}{s} = \frac{p}{q}+\frac{q-p}{q} = 1. \end{align*} Choose a measurable representative of the equivalence class $f$. Define the [measurable functions](/page/Measurable%20Functions) $h:X\to[0,\infty]$, $x\mapsto |f(x)|^p$, and $\mathbb{1}_X:X\to\mathbb{R}$, $x\mapsto 1$. Because $f\in L^q(X,\mathcal{A},\mu;\mathbb{F})$, the function $h$ belongs to $L^r(X,\mathcal{A},\mu;\mathbb{R})$, with \begin{align*} \|h\|_r = \left(\int_X |h(x)|^r\,d\mu(x)\right)^{1/r} = \left(\int_X |f(x)|^q\,d\mu(x)\right)^{p/q} = \|f\|_q^p. \end{align*} Also $\mathbb{1}_X\in L^s(X,\mathcal{A},\mu;\mathbb{R})$, since $\mu(X)<\infty$, and \begin{align*} \|\mathbb{1}_X\|_s = \left(\int_X 1\,d\mu(x)\right)^{1/s} = \mu(X)^{1/s} = \mu(X)^{(q-p)/q}. \end{align*} Applying the conjugate pairing estimate [citetheorem:9905] to $h\in L^r$ and $\mathbb{1}_X\in L^s$ gives \begin{align*} \int_X |f(x)|^p\,d\mu(x) = \int_X h(x)\mathbb{1}_X(x)\,d\mu(x) \leq \|h\|_r\|\mathbb{1}_X\|_s. \end{align*} Substituting the two norm computations yields \begin{align*} \int_X |f(x)|^p\,d\mu(x) \leq \|f\|_q^p\,\mu(X)^{(q-p)/q}. \end{align*} Since \begin{align*} \frac{q-p}{q} = p\left(\frac{1}{p}-\frac{1}{q}\right), \end{align*} we obtain \begin{align*} \|f\|_p^p \leq \mu(X)^{p(1/p-1/q)}\|f\|_q^p. \end{align*} Taking $p$-th roots gives \begin{align*} \|f\|_p \leq \mu(X)^{1/p-1/q}\|f\|_q. \end{align*} In particular $\|f\|_p<\infty$, so $f\in L^p(X,\mathcal{A},\mu;\mathbb{F})$.[/step]

[guided]Assume first that $q<\infty$. The goal is to control the $L^p$ norm of $f$ using the stronger $L^q$ information. Since the $L^p$ norm is built from the integral of $|f|^p$, we rewrite that integrand as a product: \begin{align*} |f(x)|^p = |f(x)|^p\cdot 1. \end{align*} This is exactly the form where Hölder's inequality applies. Define \begin{align*} r := \frac{q}{p}, \qquad s := \frac{q}{q-p}. \end{align*} Because $1\leq p<q<\infty$, both $r$ and $s$ lie in $(1,\infty)$, and they are conjugate exponents: \begin{align*} \frac{1}{r}+\frac{1}{s} = \frac{p}{q}+\frac{q-p}{q} = 1. \end{align*} Now define the measurable functions \begin{align*} h:X &\to [0,\infty] \end{align*} \begin{align*} x &\mapsto |f(x)|^p \end{align*} and \begin{align*} \mathbb{1}_X:X &\to \mathbb{R} \end{align*} \begin{align*} x &\mapsto 1. \end{align*} We verify the hypotheses needed to apply the conjugate pairing estimate. First, $h\in L^r$ because \begin{align*} \int_X |h(x)|^r\,d\mu(x) = \int_X |f(x)|^{pr}\,d\mu(x) = \int_X |f(x)|^q\,d\mu(x) <\infty. \end{align*} Moreover, \begin{align*} \|h\|_r = \left(\int_X |h(x)|^r\,d\mu(x)\right)^{1/r} = \left(\int_X |f(x)|^q\,d\mu(x)\right)^{p/q} = \|f\|_q^p. \end{align*} Second, $\mathbb{1}_X\in L^s$ precisely because the [measure space](/page/Measure%20Space) has finite total measure: \begin{align*} \int_X |\mathbb{1}_X(x)|^s\,d\mu(x) = \int_X 1\,d\mu(x) = \mu(X) <\infty. \end{align*} Its norm is \begin{align*} \|\mathbb{1}_X\|_s = \left(\int_X 1\,d\mu(x)\right)^{1/s} = \mu(X)^{1/s} = \mu(X)^{(q-p)/q}. \end{align*} The conjugate pairing estimate [citetheorem:9905] now applies to $h\in L^r$ and $\mathbb{1}_X\in L^s$, since $r$ and $s$ are conjugate. It gives \begin{align*} \int_X |f(x)|^p\,d\mu(x) = \int_X h(x)\mathbb{1}_X(x)\,d\mu(x) \leq \|h\|_r\|\mathbb{1}_X\|_s. \end{align*} Substituting the computed norms gives \begin{align*} \int_X |f(x)|^p\,d\mu(x) \leq \|f\|_q^p\,\mu(X)^{(q-p)/q}. \end{align*} The exponent on $\mu(X)$ has the correct form because \begin{align*} \frac{q-p}{q} = p\left(\frac{1}{p}-\frac{1}{q}\right). \end{align*} Therefore \begin{align*} \|f\|_p^p \leq \mu(X)^{p(1/p-1/q)}\|f\|_q^p. \end{align*} Taking $p$-th roots gives \begin{align*} \|f\|_p \leq \mu(X)^{1/p-1/q}\|f\|_q. \end{align*} This also proves $f\in L^p(X,\mathcal{A},\mu;\mathbb{F})$, because the right-hand side is finite.[/guided]

[step:Use the essential supremum bound when $q=\infty$] Assume now that $q=\infty$. Let $M := \|f\|_\infty$. By the definition of the $L^\infty$ norm as the essential supremum, the inequality $|f(x)|\leq M$ holds for $\mu$-almost every $x\in X$. Since $|f(x)|^p$ and the constant function $M^p$ are nonnegative measurable functions and the [Lebesgue integral](/page/Lebesgue%20Integral) respects almost-everywhere inequalities for such functions, we get \begin{align*} \int_X |f(x)|^p\,d\mu(x)\leq\int_X M^p\,d\mu(x)=M^p\mu(X). \end{align*} Thus \begin{align*} \|f\|_p^p\leq\mu(X)\|f\|_\infty^p. \end{align*} Taking $p$-th roots gives \begin{align*} \|f\|_p\leq\mu(X)^{1/p}\|f\|_\infty. \end{align*} Since $1/q=0$ when $q=\infty$, this is exactly \begin{align*} \|f\|_p\leq\mu(X)^{1/p-1/q}\|f\|_q. \end{align*} Again the right-hand side is finite, so $f\in L^p(X,\mathcal{A},\mu;\mathbb{F})$. [/step]

[step:Record the zero-measure case and complete the inclusion] The preceding estimates remain valid when $\mu(X)=0$. Indeed, for finite $q$, the integral formula gives \begin{align*} \|f\|_p^p=\int_X |f(x)|^p\,d\mu(x)=0. \end{align*} and for $q=\infty$ the essential supremum norm is $0$ for every equivalence class on a null measure space. Thus the displayed inequality holds with the factor $\mu(X)^{1/p-1/q}$, whose exponent is positive because $p<q\leq\infty$. Combining the finite-$q$ and $q=\infty$ cases proves that every $f\in L^q(X,\mathcal{A},\mu;\mathbb{F})$ belongs to $L^p(X,\mathcal{A},\mu;\mathbb{F})$ and satisfies \begin{align*} \|f\|_p\leq\mu(X)^{1/p-1/q}\|f\|_q. \end{align*} This proves the claimed continuous inclusion. [/step]