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*}
 A: 
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 $E(Y)=2$.
A: As Did says, this is false.
Suppose that $I=\{0,1\}$ and $X(u)$ and $\theta$ are determined by a fair coin toss:
$$
\begin{align}
\text{heads}&\implies X(0)=2, X(1)=0, \theta=0\\
\text{tails}&\implies X(0)=0, X(1)=2, \theta=1
\end{align}
$$
Then $E(X(0))=E(X(1))=1$ and $E(X(\theta))=2$.
A: 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.
