For a sequence of nonnegative and absolutely continuous distribution functions $F_{n}$ with density $f_{n}$, $n = 1,2, \cdots$, $$\lim_{n \to \infty} \int_{0}^{\infty} xdF_{n}(x) = \lim_{n \to \infty} \int_{0}^{\infty} x f_{n}(x) dx, \, \text{which is not guaranteed to be equal to} \\ \int_{0}^{\infty}x \lim_{n \to \infty}f_{n}(x)dx $$
I need to come up with two examples illustrating that $\lim_{n\to \infty}\int_{0}^{\infty} x f_{n}(x)dx$ is not necessarily equal to $\int_{0}^{\infty}x\lim_{n \to \infty}f_{n}(x)dx$.
Thus far, I have the example where $f_{n}(x) = n^{3}I_{(0,\frac{1}{n})}(x)$:
Fix $x_{0} \in (0,1)$. Then, for $n \geq N$, $f_{n}(x_{0})=n^{3} \cdot 0 = 0$, and since $x_{0} $ was arbitrary, we have that $\lim_{n \to \infty} f_{n}(x) = 0$.
Therefore, $\int_{0}^{\infty}x\lim_{n \to \infty}f_{n}(x)dx = \int_{0}^{\infty}x \cdot 0 dx = 0$.
However, $\displaystyle \lim_{n \to \infty} \int_{0}^{\infty} xf_{n}(x)dx = \int_{0}^{\infty}xn^{3}\cdot 1_{(0, \frac{1}{n})}(x) dx = \lim_{n \to \infty}n^{3} \int_{0}^{\frac{1}{n}}xdx = \lim_{n \to \infty}n^{3}\left[ \frac{x^{2}}{2}\right]_{0}^{1/n} = \lim_{n \to \infty}\left(n^{3}\left[\left(\frac{1}{2}\right)\left(\frac{1}{n^2}\right)\right]\right)= \lim_{n \to \infty}\frac{n}{2} = \infty$
So, this seems to work, although I don't know how I can reframe it in terms of a sequence of nonnegative and absolutely continuous distribution functions $F_{n}$ with density $f_{n}$...
Can I turn it into this by letting $\Omega = [0,1]$, $\mathcal{F} = \mathcal{B}([0,1])$ (where $\mathcal{B}([0,1])$ denotes the Borel sigma field over the interval $[0,1]$), and $P(dx) = dx$?
If I can do this, then I still need one more example, although all I seem to be able to come up with are variations on this one, with different powers of $n$ inside and outside of the indicator function.
Could anybody let me know of another one, please?
Thank you for your time and patience.
A comment: it looks to me like you intend these to be probability density functions. If so, you've got a small issue: a probability density function $f(x)$ requires that $\int_{-\infty}^{\infty} f(x) \, \textrm{d} x = 1$; your density functions don't satisfy this. (If I'm wrong in my assumption, please feel free to ignore this line of commentary!)
Otherwise, the question of how to reframe this in terms of $F_n$ and $f_n$ isn't too onerous; you can just define a valid density function on the real number line (make it so that $f(x) \geq 0$ and that it satisfies the integration condition above), and then set $F(x) = \int_{-\infty}^x f(t)\, \textrm{d}t$, which is then automatically absolutely continuous. Your sample space then just becomes $\mathbb{R}$, the $\sigma$-field can be the Lebesgue-measurable sets (or Borel-measurable; it almost surely won't matter), and the probability measure is then defined by $\mathbb{P}(A) = \int_A f(x) \, \textrm{d} x.$ In other words: these details aren't the ones to worry about. Your task is to worry about the density functions doing the thing they're supposed to do.
So, back to the densities: if I'm right and your density functions don't work (because you're supposed to be talking probability here), you still have a good kernel of an idea: consider functions that are nonzero constants on some intervals. Instead of having the interval shrink and the function value increase, try the reverse. (This is motivated by leveraging that $x$ in your integral, which will help you out if $x$ is actually allowed to become big in your integral.)
For a different type of example, consider functions that are $1$ on an interval of unit length and are $0$ elsewhere, but have the interval "move" as $n$ increases. There are many other types of examples, but these were the first two that came to my mind.