[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 $f\in L^p(X,\mathcal{A},\mu;\mathbb{F})$.
[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]
[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]
