Theorem 10.1.1 in Prekopa establishes the following result.
Suppose $h:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}$ is continuous and $\xi$ is a random vector from a probability space $(\Xi,\mathcal{F},\mathbb{P})$ with $\Xi \subset \mathbb{R}^m$ and $\mathbb{P}\{h(x,\xi) = 0\} = 0$, $\forall x \in \mathbb{R}^n$. Let $\mathcal{i}$ denote the indicator function of the positive reals, i.e. $$\mathcal{i}(z) := \begin{cases} 1 &\mbox{if } z > 0 \\ 0 & \mbox{if } z \leq 0\end{cases}.$$
Then $\mathbb{E}\left[\mathcal{i}(h(x,\xi))\right]$ is a continuous function of $x$, where $\mathbb{E}$ denotes the expectation operator.
My question is the following: Let $g:\mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}$ be a continuous function such that $\left\lvert g(x,\xi) \right\rvert < M$ for some $M > 0$ (this ensures that the next expectation is well-defined). Do the above conditions also guarantee that $\mathbb{E}\left[\left\lvert\mathcal{i}(h(x,\xi)) - g(x,\xi)\right\rvert\right]$ is a continuous function of $x$, or do we need to impose additional assumptions?
Clearly, we have that the above expectation is continuous if the absolute value is omitted. I have a feeling that continuity still holds even if we include the absolute value. I've tried to use dominated convergence to show this - in the process, I've had to show that for each sequence $\{x_n\} \to x$ and $\xi \in \Xi$, $\underset{n \to \infty}{\lim} \mathcal{i}(h(x_n,\xi))$ converges to $\mathcal{i}(h(x,\xi))$ in the $L^1$-norm, which eludes me.