Upper bound on the expectation of a random variable indexed by a random variale Imagine that we are given $X(u)$, a random variable indexed by some parameter $u \in I \subset \mathbb{R}$. Furthermore, assume that we know that:
\begin{equation*}
\mathbb{E}(|X(u)|)\leq 1 \quad \text{$\forall$ $u\in I$.}
\end{equation*}
Now, let $\theta$ be a random variable such that $\mathbb{P}(\theta \in I) = 1$. We do not suppose that $\theta$ and $X(u)$ are independent.
Does it hold that:
\begin{equation*}
\mathbb{E}(|X(\theta)|)\leq 1 \quad \text{?}
\end{equation*}
At least one declared downvote for extramathematical reasons, by user @stuartstevenson. This sucks.


The result does not hold without independence, as the simple example below shows.

Consider the parameter space $I=[0,1)$ and the probability space $(\Omega,\mathcal F,P)$ defined by $\Omega=[0,1)$, $\mathcal F=\mathcal B(\Omega)$, and $P$ the Lebesgue measure restricted to $\Omega$.

For every $u$ in $I$ and $\omega$ in $\Omega$, let $X(u)(\omega)=2\cos^2(2\pi(\omega+u))$. For every $\omega$ in $\Omega$, let $\theta(\omega)=1-\omega$.

Then, $E(X(u))=1$ for every $u$ in $I$ but the random variable $Y=X(\theta)$ is, by definition, such that $Y(\omega)=X(1-\omega)(\omega)=2$ for every $\omega$ in $\Omega$ hence $E(Y)=2$.
Yes, it holds.

If $\theta \in I$ with probability $1$ then $\theta \in I$.

The rest follows from definition. 

The only issue I might see is if this is in some kind of limiting sense in which case you might use the suffix, "in the limit" or "almost surely" depending on the context.