Let $f(x,y,\theta)$ be a real-valued function of $x \in X$ and $y \in Y$, and let's interpret $\theta \in \Theta$ as a parameter. Let $X$, $Y$ and $\Theta$ be closed intervals of the real line. Then, $f: X \times Y \times \Theta \to \mathbb{R}$. The function is bounded: $f(x,y,\theta)<\infty$ always.
Fix some $\theta \in \Theta$ and some $y \in Y$. The function $f(x,y,\theta)$ is continuous in $x$ for all $x \in X$, concave ($\frac{\partial^2 f(x,y,\theta)}{\partial x^2}\leq 0$), twice differentiable in all $x \in X$, and therefore single-peaked. Moreover, for all $x \in X$ and $y \in Y$, the function $f(x,y, \theta)$ is continuos in $\theta$ for all $\theta \in \Theta$.
Let $ y \sim \eta$, where $\eta$ is some probability density function: $\eta: Y \to [0,1]$. I want to show that the expectation function: \begin{equation} \mathbb{E}_\eta[f(x,y,\theta)]=\int f(x,y,\theta) d \eta \end{equation}
is continuous in $\theta$ for all $\theta \in \Theta$.
Any hint on how to proceed? Many thanks for considering my request.
[Edit: I've removed a uniform continuity argument as it was not needed for the proof, and the result would actually hold just as well on an open interval $\Theta$ on which $f$ was not uniformly continuous.]
We only need continuity of $f$ in $\theta$ and boundedness of $f$ for the conclusion. None of the conditions on how the function behaves in $x$ are relevant.
Here is an outline of the argument.
Since $f$ is bounded, suppose $f\leq M$ everywhere.
Fix $x\in X$, $\theta\in \Theta$, and fix $\epsilon>0$. For each $y\in Y$, continuity of $f$ at $\theta$ in $\Theta$ implies there is a $\delta_y>0$ such that $f|(x,y,\theta)-f(x,y,\theta')|<\epsilon$ whenever $|\theta-\theta'|<\delta_y$.
Define $Y_\delta=\{y\in Y\mid \delta_y\geq\delta\}$. Then $Y$ is an increasing union $$Y=\bigcup_{\delta>0} Y_\delta\text{.}$$ Therefore we may pick $\delta>0$ so that $\eta(Y\backslash Y_\delta)$ is as small as we like.
Now for $|\theta-\theta'|<\delta$, estimate
$$\int_Y \left| f(x,y,\theta)-f(x,y,\theta')\right| d\eta$$
from above by splitting the integral up between $Y_\delta$ and $Y\backslash Y_\delta$.
I will leave as an exercise to compute how small $\eta(Y\backslash Y_\delta)$ must be in order to get the estimate to converge.
Remark
Also we should in principle be cautious about measurability of the sets $Y_\delta$... but this can be taken care of by defining $\delta_y$ as a supremum of legal choices of $\delta_y$. I am assuming that $f$ is reasonably well behaved, e.g., Borel measurable.