Rigorous justification for this conditional probability inequality? Let $(\Omega,\mathcal F,P)$ be a probability space. Consider a collection of bounded real random variables $X(\gamma)$, for $\gamma\in[0,1]$, defined on this probability space. Let $(\gamma_i)_{i=1}^\infty$ be a sequence of iid random variables taking values in $[0,1]$. The family $\{\gamma_i : i \in \mathbb{N}\}$ is assumed to be independent of the family $\{X(\gamma) : \gamma \in [0,1]\}$. Consider the following inequality for a given $\delta>0$:
$$\sup_{i\in\mathbb{N}}P\bigg(\Big|X(\gamma_i)-E[X(\gamma_i)|\sigma(\gamma_i)]\Big|>\delta \bigg | \sigma(\gamma_i) \bigg)\leq \sup_{\gamma \in [0,1]}P\bigg(\Big|X(\gamma)-E[X(\gamma)]\Big|>\delta \bigg) .$$
It seems to me that this inequality is true, since once we condition on $\gamma_i$, the $i$th probability on the left must appear on the right as well.
But how to show it rigorously? Any help on this is very appreciated.
EDIT: In fact I would be happy if someone could just give a rigorous proof of the statement
$$E[X(\gamma_i)|\sigma(\gamma_i)(\omega)=E[X(\gamma_i(\omega)) ,\quad \omega\in\Omega.$$
 A: Okay, so here are my few cents. I hope that others point out my mistakes, which I almost definitely made.
Let's take the left side appart:
For each $i\in\mathbb{N}$, the conditional expectation
$$E[X(\gamma_i)|\sigma(\gamma_i)]:\Omega\to\mathbb{R}$$
is a random variable in the sense that
$$E[X(\gamma_i)|\sigma(\gamma_i)](\omega):=E[X(\gamma_i(\omega))],\quad \omega\in\Omega.$$
By construction of the conditional expactation, this function factorizes over $\gamma_i$, i.e. there exists a function $h:[0,1]\to \mathbb{R}$ measurable, such that
$$E[X(\gamma_i)|\sigma(\gamma_i)](\omega) = h\circ \gamma_i(\omega),\quad \omega\in\Omega.$$
For this, we need the measurability condition from above.
In the same way - "conitional probability is just conditional expectation of an indicator function" - the conditional probability
$$ P\bigg(\Big|X(\gamma_i)-E[X(\gamma_i)|\sigma(\gamma_i)]\Big|>\delta \bigg | \sigma(\gamma_i) \bigg):\Omega\to\mathbb{R}$$
is also "just" a measurable function in the sense that
$$P\bigg(\Big|X(\gamma_i)-E[X(\gamma_i)|\sigma(\gamma_i)]\Big|>\delta \bigg | \sigma(\gamma_i) \bigg)(\omega)=P\bigg(\Big|X(\gamma_i(\omega))-E[X(\gamma_i(\omega))]\Big|>\delta \bigg)=g\circ \gamma_i(\omega)$$
for a measurable $g:[0,1]\to [0,1]$.
Bringing both sides together:
On the left, we have
$$\sup_{i\in\mathbb{N}}g\circ \gamma_i:\Omega\to\mathbb{R}.$$
For each $i\in\mathbb{N}$ and $\omega\in\Omega$, we indeed have
$$g( \gamma_i(\omega))\leq \sup_{\gamma\in[0,1]}g(\gamma)=\sup_{\gamma\in[0,1]} P\bigg(\Big|X(\gamma)-E[X(\gamma)]\Big|>\delta \bigg).$$
Taking the supremum on the left side yields the desired inequality.
Notes:

*

*I still don't see, where the independence of $(X(\gamma))_{\gamma\in[0,1]}$ and $(\gamma_i)_{i\in\mathbb{N}}$ is needed. Maybe @chris-janjigian could help and spot my error of reasoning.

*@chris-janjigian provided a source above. As we both noted, one needs the measurability of $(\gamma,\omega)\mapsto X(\gamma)(\omega):=X(\gamma(\omega))(\omega)$.

*I don't think that one needs the $(\gamma_i)_{i\in\mathbb{N}}$ to be iid.

A: This doesn't answer the original question, but here is an argument for the case where $X(\cdot)$ is a random variable taking values in $C[0,1]$. The same idea works if it is either right- or left- continuous. For simplicity, I will keep the boundedness hypothesis and without loss of generality, take $\|X\|_\infty \leq 1$. It's not needed, but will simplify things a bit.
Just to sketch out the main idea, if $X$ is continuous, you can pretend $\gamma_i$ is a random variable taking only finitely many possible values by approximating it by one. This is always possible by a truncation argument. Path continuity lets you undo the discretization by taking a limit. There may be an abstract way of doing this more generally, but I couldn't see the argument quickly. I probably need to spend some time refreshing myself on measure theory.
Anyway, the first claim is that if we define $g(\gamma) = E[X(\gamma)] = \int X(\gamma,\widetilde{\omega}) P(d\widetilde{\omega})$ then $E[X(\gamma_i)|\gamma_i](\omega) = g(\gamma_i(\omega))$ almost surely. Be careful about the notation here, it is not true that $g(\gamma_i) = E[X(\gamma_i)]$.  The former is a random variable and the second is a constant. Because of this notational issue, I am going to keep the integrals around, even though it looks ugly. The claim is that for all $h \in C_b[0,1]$,
$$
\int X(\gamma_i(\omega),\omega) h(\gamma_i(\omega)) P(d\omega) = \int \int X(\gamma_i(\omega),\widetilde{\omega})P(d\widetilde{\omega}) h(\gamma_i(\omega)) P(d\omega).
$$
Call $\eta_m(\omega) = \sum_{k=0}^{m-1} \frac{k}{m} 1_{[k/m,(k+1)/m)}(\gamma_i(\omega)) + 1_{\{1\}}(\gamma_i(\omega))$. Then we have $\lim_{m\to\infty} \eta_m(\omega) = \gamma_i(\omega)$ surely. Note also that $\eta_m$ takes on only finitely many values. By dominated convergence,
$$
\lim_{m\to\infty}E[X(\eta_m)h(\eta_m)] = \lim_{m\to\infty} \int X(\eta_m(\omega),\omega) h(\eta_m(\omega)) P(d\omega) \\=\int X(\gamma_i(\omega),\omega) h(\gamma_i(\omega)) P(d\omega) = E[X(\gamma_i)h(\gamma_i)].
$$
We used path regularity and smoothness of our test function in this step to obtain the almost sure convergence of the integrand. As a comment here, by playing around with the form of the truncation, it would suffice to assume that $X$ is either right- or left- continuous.
On the other hand,
$$E[X(\eta_m)h(\eta_m)] = \sum_{k=0}^m E\left[X(\eta_m)h(\eta_m) 1_{\{\eta_m = k/m\}}\right] \\= \sum_{k=0}^m E\left[X(k/m)h(k/m) 1_{\{\eta_m = k/m\}}\right] \\= \sum_{k=0}^m E[X(k/m)]h(k/m)P(\eta_m=k/m) \\= \int \int X(\eta_m(\omega),\widetilde{\omega}) P(d\widetilde{\omega})h(\eta_m(\omega)) P(d\omega)$$
In the third equality, we used independence of the process $X$ and $\sigma(\gamma_i)$. The fourth equality is the definition. Sending $m \to \infty$ and again using dominated convergence, we have
$$
\int X(\gamma_i(\omega),\omega) h(\gamma_i(\omega)) P(d\omega) = \int \int X(\gamma_i(\omega),\widetilde{\omega})P(d\widetilde{\omega}) h(\gamma_i(\omega)) P(d\omega).
$$
This gives the claim.
To answer your question (with the extra hypothesis anyway), it suffices to show that if we define
$$q(\gamma,\delta) = P(|X(\gamma) - E[X(\gamma)]|>\delta) = \int 1_{\{|X(\gamma,\widetilde{\omega}) - g(\gamma)|>\delta\}}P(d\widetilde{\omega})$$
then $P(|X(\gamma_i) - E[X(\gamma_i)|\gamma_i] > \delta|\gamma_i) = q(\gamma_i(\omega),\delta)$ almost surely. We have shown that we can replace $E[X(\gamma_i)|\gamma_i]$ in this expression with $g(\gamma_i)$.
Let $f \in C_b([-1,1] \times [0,1])$ be given. The same argument as above shows that for $h \in C_b(0,1)$,
$$
\lim_{m\to\infty}E[f(X(\eta_m),\eta_m)h(\eta_m)] = E[f(X(\gamma_i),\gamma_i)h(\gamma_i)]
$$
and
$$
E[f(X(\eta_m),\eta_m)h(\eta_m)] = \int\int f(X(\eta_m(\omega),\widetilde{\omega}),\eta_m(\omega))P(d\widetilde{\omega}) h(\eta_m(\omega)) P(d\omega)
$$
Sending $m \to \infty$ lets us replace $\eta_m$ by $\gamma_i$ in this expression. By density, this holds for all bounded Borel functions in place of $f$ and $h$ and in particular in the case where $f(x,y) = 1_{\{|x-g(y)|>\delta\}}$. We then have for any $h \in B_b[0,1]$
$$
E[f(X(\gamma_i),\gamma_i)h(\gamma_i)] =  \int\int f(X(\gamma_i(\omega),\widetilde{\omega}),\gamma_i(\omega))P(d\widetilde{\omega}) h(\gamma_i(\omega)) P(d\omega) \\=  \int q(\gamma_i(\omega),\delta) h(\gamma_i(\omega)) P(d\omega),
$$
which is the desired result.
A: Okay, in my opinion, the crux of the problem is to make sense of all the symbols written up there.
I'm almost there but I came up with two questions I could not yet answer:
Fix $i\in\mathbb{N}$. What is the object $X(\gamma_i)$? I thought it to be the function
$$X(\gamma_i):\Omega \to \mathbb{R},\quad X(\gamma_i)(\omega):=X(\gamma_i(\omega))(\omega).$$
If so, why is this even a random variable, i.e. why is it measurable? Do we have other assumptions on the concrete distributions of the $\gamma_i$, for example they take values in a discrete subset of $[0,1]$?
I'm sorry, I don't have the reputation to make a comment yet, so I had to write an "answer".
