Proof that Muckenhoupt's $A_q$ Condition Implies $A_p$ for $p<q$?

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It is said $f\in A_p$ if it satisfies the following (Muckenhoupt's $A_p$) condition: $$\underset{I\subset\mathbb{R}}\sup\left(\dfrac{1}{|I|}\int_If(t)dt\right)\left(\dfrac{1}{|I|}\int_If(t)^{-\frac{p'}{p}}\right)^\frac{p}{p'}<\infty$$ where the $\sup$ is taken over bounded intervals $I$ in $\mathbb{R}$, and $p'$ is the Holder conjugate of $p$.

I find the following fact that $A_1\subset A_p\subset A_\infty$ for $1\leq p\leq\infty$ on Wikipedia's page, but have not found this proven in any papers so far (nor have found a proof myself).

Any reference to this would be appreciated.

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For any nonnegative measurable function $h$ on a probability space $\Omega$ the quasi-norm $\|h\|_q=\left(\int_{\Omega} h^q\right)^{1/q}$ is increasing with $q$, $0<q<\infty$. Indeed, Jensen's inequality gives $$ \left(\int_{\Omega} h\right)^q \le \int_{\Omega} h^q \tag1 $$ which says $\|h\|_1\le \|h\|_q$ for $q\ge 1$. Given $0< q_1<q_2<\infty$, apply (1) to $h^{q_1}$ with $q=q_2/q_1$.

Specialize the above to $\Omega$ being an interval $I$ with normalized Lebesgue measure, and $h=1/f$. Then $$ \left(\dfrac{1}{|I|}\int_If(t)^{-\frac{p'}{p}}\right)^\frac{p}{p'} = \|h\|_q \tag2$$ where $q=p'/p$ is a decreasing function of $p$. Therefore, (2) is a decreasing function of $p$, which implies that the inequality $$\underset{I\subset\mathbb{R}}\sup\left(\dfrac{1}{|I|}\int_If(t)dt\right)\left(\dfrac{1}{|I|}\int_If(t)^{-\frac{p'}{p}}\right)^\frac{p}{p'}<\infty$$ becomes weaker (more permissive) as $p$ increases.

To handle $A_\infty$, one can introduce $\|h\|_0=\exp\int_\Omega \log h$ and argue that $\|h\|_0\le \|h\|_q$ for all $q>0$; the rest goes as above.

A reference is Harmonic Analysis by Stein, section V.1.3, page 195:

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