Proving conditional Hölder inequality using regular conditional distribution I am trying to prove the conditional Hölder inequality using regular conditional distributions. The inequality I'm trying to prove is:

For $p,q \in (1,\infty)$ with $\frac 1 p + \frac 1 q = 1$, and for $X \in \mathcal L^p(\Omega, \mathcal A, \mathbb P)$ and $Y \in L^q(\Omega, \mathcal A, \mathbb P)$, and for $\mathcal F \subset \mathcal A$ a sub-$\sigma$-algebra, almost surely we have
$$
\mathbb E \left[ |XY|\,\big|\,\mathcal F\right] \leq \mathbb E\left[|X|^p\,\big|\,\mathcal F\right]^{1/p}\mathbb E\left[|Y|^q\,\big|\,\mathcal F\right]^{1/q}
$$

I've found lots of proofs of this fact, but I specifically am trying to prove it using a theorem of regular conditional distributions:

Let $X$ be a random variable on $(\Omega, \mathcal A, \mathbb P)$ with values in a Borel space $(E,\mathcal E)$, $\mathcal F \subset \mathcal A$ is a sub-$\sigma$-algebra, and $\kappa_{X,\mathcal F}$ a regular conditional distribution of $X$ given $\mathcal F$. Further, let $f : E \to \mathbb R$ be measurable and $\mathbb E[|f(x)|] < \infty$. Then,
$$
\mathbb E\left[f(x)\,|\,\mathcal F\right](\omega) = \int_E f(x)\kappa_{X,\mathcal F}(\omega, dx) \quad \textrm{for $\mathbb P$-almost all $\omega\in\Omega$}.
$$

Applying Young's inequality and monotonicity and linearity of conditional expectation gives me
$$
\mathbb E \left[ |XY|\,\big|\,\mathcal F\right](\omega) \leq \frac 1 p \mathbb E\left[|X|^p\,\big|\,\mathcal F\right](\omega) + \frac 1 q \mathbb E\left[|Y|^q\,\big|\,\mathcal F\right](\omega) = \frac 1 p \int |x|^p\kappa_{X,\mathcal F}(\omega,dx) + \frac 1 q \int |y|^q\kappa_{Y,\mathcal F}(\omega,dy)
$$
but I'm having trouble getting from here to the desired inequality. Alternatively, the standard Hölder's inequality gives us $\mathbb E\left[|XY|\right]<\infty$, so the above result also implies
$$
\mathbb E \left[ |XY|\,\big|\,\mathcal F\right](\omega) = \int_{\mathbb R^2}|xy| \kappa_{X \times Y,\mathcal F}(\omega, dx dy)
$$
But both of these approaches have led me to circular arguments or using measures that I don't think formally exist (like $A \mapsto \mathbb P[A|\mathcal F](\omega)$ for a fixed $\omega\in\Omega$). Any suggestions or other places to look?
 A: Again, I am not answering your question.  But this is too large for the comments.
When proving the standard Holder's inequality, we actually use Young's inequality in this form: for any $x,y \ge 0$, $\lambda > 0$
$$ xy \le (\lambda x) (\lambda^{-1} y) \le \tfrac1p \lambda^p x^p + \tfrac1q \lambda^{-q} y^q $$
from which you get
$$ E(|XY|) \le \tfrac1p \lambda^p E(|X|^p) + \tfrac1q \lambda^{-q} E(|Y^q|) . $$
Then you use: if $A,B \ge 0$:
$$ \inf_{\lambda >0} \left(\tfrac1p \lambda^p A^p + \tfrac1q \lambda^{-q} B^q\right) = AB. $$
(This is just putting the conditions for equality into Young's inequality.)
In proving the conditional form of Holder's inequality, the infimum will be taken over $\lambda$ a positive $\mathcal F$-measurable function.
But what this does say is that if you do want to use conditional regular distributions, you really should be using the form of Young's inequality I wrote above.
A: Let $\pi_1, \pi_2 : \mathbb R^2 \to \mathbb R$ be the projections $\pi_1(x,y) = x$ and $\pi_2(x,y) = y$. After showing $\kappa_{X,\mathcal F}(\omega,\cdot) = (\pi_1)_*\kappa_{(X,Y),\mathcal F}(\omega,\cdot)$,
$$
\int_{\mathbb R^2}|x|^p\kappa_{(X,Y),\mathcal F}(\omega, dx dy) = \int_{\mathbb R} |x|^p \kappa_{X,\mathcal F}(\omega, dx) = \mathbb E\left[ |X|^p\,\big|\,\mathcal F\right](\omega)
$$
by the cited result on regular conditional distributions, which is finite for a.e. $\omega\in\Omega$. So $|\pi_1| \in \mathcal L^p\left(\mathbb R^2, \mathcal B(\mathbb R^2), \kappa_{(X,Y),\mathcal F}(\omega,\cdot)\right)$, and similarly $|\pi_2| \in \mathcal L^q\left(\mathbb R^2, \mathcal B(\mathbb R^2), \kappa_{(X,Y),\mathcal F}(\omega,\cdot)\right)$, for a.e. $\omega\in\Omega$.
So,
\begin{align*}
\mathbb E\left[|XY|\,\big|\,\mathcal F\right](\omega) &= \int_{\mathbb R^2} |xy| \kappa_{(X,Y),\mathcal F}(\omega, dxdy) \\ &\qquad\qquad\qquad\textrm{by the cited result on regular conditional distributions;} \\
&\leq \left(\int_{\mathbb R^2} |x|^p \kappa_{(X,Y),\mathcal F}(\omega, dxdy)\right)^{1/p}\left(\int_{\mathbb R^2} |y|^q \kappa_{(X,Y),\mathcal F}(\omega, dxdy)\right)^{1/q} \\
&\qquad\qquad\qquad\textrm{by the standard Hölder's inequality applied to }\left(\mathbb R^2, \kappa_{(X,Y),\mathcal F}(\omega,\cdot)\right); \\
&= \mathbb E\left[|X|^p\,\big|\,\mathcal F\right]^{1/p}(\omega)\mathbb E\left[|Y|^q\,\big|\,\mathcal F\right]^{1/q}(\omega)\\
&\qquad\qquad\qquad\textrm{by the cited result and using the image measure properties of $\kappa_{X,\mathcal F}$ and $\kappa_{Y,\mathcal F}$.}
\end{align*}
A: How about starting with
$$\mathbb E \left[\frac{|X|}{\mathbb E[|X|^p|\mathcal F]^{1/p}} \frac{|Y|}{\mathbb E[|Y|^q|\mathcal F]^{1/q}} \Bigg | \mathcal F \right] ?$$
If $Z$ is $\mathcal F$ measurable, then
$$ \mathbb E(f(X) Z | \mathcal F)(\omega) = Z(\omega) \mathbb E(f(X) | \mathcal F)(\omega)
= Z(\omega) \int_E f(x) \kappa_{X,\mathcal F}(\omega,dx) .$$
To avoid zero and infinity issues, first apply it to $X_{\epsilon,N} = (|X| \vee \epsilon )\wedge N$, and similarly for $Y$, and then let $\epsilon \to 0+$, and $N \to \infty$.
Of course, when you do the Young's inequality at the beginning, the introduction of the regular conditional distribution is an extra step that serves no purpose.
