How to show this $\phi$ mixing inequality? Let $(\Omega,\mathcal A,P)$ be a probability space and $\mathcal F$ a sub-$\sigma$-algebra of $\mathcal A$. Am trying to show that
$$ |P(A|\mathcal F)-P(A)|\leq \phi(\mathcal A,\mathcal F) \quad \quad P\text{-almost surely}$$
for all $A\in \mathcal A$, where $\phi(\mathcal A,\mathcal F)=\sup_{A\in \mathcal A, F\in \mathcal F, P(F)>0} \bigg|P(A|F)-P(A)\bigg|$.
Seems true but I don't see where to start. Any ideas on how to proceed?
 A: Lemma. Let $X$ be a real random variable on the probability space $(\Omega,\mathcal A,P)$ with $\|X\|_{\infty}<\infty$. Suppose that $X$ is measurable with respect to the sub-$\sigma$-algebra $\mathcal F$ of $\mathcal A$. Then
$$\sup_{F\in\mathcal F,P(F)>0}\frac{1}{P(F)}\bigg|\int_F X dP\bigg|=\|X\|_{\infty} \quad \quad \quad (1)$$
Proof. If $M\geq 0$ is such that $|X|\leq M$ $P$-almost surely, and $F\in\mathcal F$ is such that $P(F)>0$, then
$$\frac{1}{P(F)}\bigg|\int_F X dP\bigg|\leq \frac{1}{P(F)}\int_F \big|X \big|dP \leq M$$
so the RHS of $(1)$ is at least as large as the LHS. The reserve inequality will obtain if we show that $$LHS\geq RHS-\epsilon$$ for all $\epsilon>0$. So let $\epsilon>0$ be given and assume WLOG that $\|X\|_{\infty}-\epsilon>0$. Define
$$F^+=\{X\geq \|X\|_{\infty}-\epsilon\}$$
$$F^-=\{-X\geq \|X\|_{\infty}-\epsilon\}$$
Note that $F^+,F^-$ are both in $\mathcal F$. Moreover at least one of them has positive probability, for otherwise this would contradict the definition of $\|X\|_{\infty}$. Let
$$F=\begin{cases}
F^+,& \text{if } P(F^+)\geq  P(F^-) \\
F^- &  \text{otherwise}
\end{cases}$$
Then $F\in\mathcal F$, $P(F)>0$, and a case check shows that
$$\frac{1}{P(F)}\bigg|\int_F X dP \bigg| \geq \|X\|_{\infty}-\epsilon $$
As $\epsilon>0$ was arbitrary the claim follows.

Now back to the original problem. Fix $A\in\mathcal A$, and let $X:=P(A| \mathcal F)-P(A)$. Note that $X$ is $\mathcal F$ measurable and $P$-almost surely bounded. Moreover if $F\in\mathcal F$ with $P(F)>0$ then
$$\frac{1}{P(F)}\bigg|\int_F X dP\bigg|=\frac{1}{P(F)} \bigg|P(A\cap F)-P(A)P(F)\bigg|=\bigg|P(A|F)-P(A)\bigg|$$
It follows from the Lemma that $$\|X\|_{\infty}=\sup_{F\in\mathcal F,P(F)>0}\frac{1}{P(F)}\bigg|\int_F X dP\bigg|\leq \phi(\mathcal A,\mathcal F)$$
and since $|X|\leq \|X\|_{\infty}$ $P$-almost surely, we are done.
