Continuity of the location Consider $f(x,\theta),$ a density function for the random variable $X$ with a parameter $\theta$. Suppose $f(x, \theta)$ is continuous in $\theta$.
Is the location
$$\int_\mathbb{R} x f(x, \theta) dx$$
continuous in $\theta$?

The expectation does not have to be a continuous function of the parameter.  The following example illustrates what can go wrong.

For any number $0 \le p \le 1,$ let $g(x,p)=0$ when $x \le 1$ and otherwise
$$g(x,p) = (p+1)x^{-(p+2)} = \frac{\mathrm{d}}{\mathrm{d}x}\left(1 - x^{-(p+1)} \right) =  \frac{\mathrm{d}}{\mathrm{d}x}G(x,p).$$
This describes a family of positive distributions with distribution functions $x\to G(x,p)$ (Pareto distributions).  Moreover, for any fixed real number $x$, the function $p\to g(x,p)$ is continuous (it's actually differentiable).
Using these, construct a family by mixing two such distributions,
$$f(x,\theta) = \frac{1 - \theta}{2} g(-x,\theta^2) + \frac{1+\theta}{2} g(x,\theta^2),$$
where $-1 \le \theta \le 1.$  Because $g$ is continuous in its second variable, the functions $\theta\to f(x,\theta)$ are all continuous (actually, differentiable).

The parameter $\theta$ plays two roles in this family: as it grows closer to $0,$ it increases the heaviness of the tails, causing the absolute value of the expectation to increase; and it also determines the amounts of negative and positive parts in the mixture, causing the expectation to move from negative to positive as $\theta$ crosses $0.$
Their expectations are
$$\begin{aligned}
e(\theta) &= \int_{\mathbb R} x f(x,\theta)\,\mathrm{d}x\\
&= \frac{1 - \theta}{2}\int_{-\infty}^0 x g(-x,\theta^2)\,\mathrm{d}x  + \frac{1 + \theta}{2}\int_0^\infty x g(x,\theta^2)\,\mathrm{d}x\\
&=  -\frac{1 - \theta}{2}\int_0^\infty x g(x,\theta^2)\,\mathrm{d}x  + \frac{1 + \theta}{2}\int_0^\infty x g(x,\theta^2)\,\mathrm{d}x\\
&=\theta \int_0^\infty x g(x,\theta^2)\,\mathrm{d}x\\
&= \theta \int_1^\infty x (\theta^2+1)x^{-(\theta^2 + 2)}\,\mathrm{d}x\\
&= \theta\left(1 + \frac{1}{\theta^2}\right) = \theta + \frac{1}{\theta},
\end{aligned}$$
provided $\theta\ne 0.$  When $\theta = 0,$ the expectation is undefined.
Clearly, as $\theta$ increases from just below $0$ to just above $0,$ there is no way to define $e(0)$ to make this a continuous function.

If you would like an example where all the density functions are continuous in $x$ everywhere (which these are not: they are discontinuous at $\pm 1$), then convolve these with (say) a standard Normal distribution.  (These will be the densities of $X+Z$ where $Z$ is an independent standard Normal variable.) That adds $0$ to all the expectations while guaranteeing all the densities are infinitely differentiable.  This is what some of them look like:

