I believed I have solved the following problem by coming up with a counterexample. I would just like some feedback on whether I have made any errors. The problem is,
Let {$F_{n}, n \geq 1$} be a sequence of distribution functions such that $F_{n}$ converges weakly to $F_{0}$. Let $g: (-\infty,\infty) \rightarrow (-\infty,\infty)$ be a continuous function. Suppose that
$\int_{-\infty}^{\infty} g dF_{n}$ exists and if finite for all $n \geq 0$
and that
$\lim_{n \to \infty} \int_{-\infty}^{\infty} g dF_{n} = \int_{-\infty}^{\infty} g dF_{0}$
Can it be concluded that g is uniformly integrable relative to $\{F_{n}, n \geq 1\}$? Give a proof or counterexample.
Here is my counterexample,
Let $P(X_{n} = -n) = P(X_{n} = n) = \frac{1}{2n}$ and $P(X_{n} = 0) = 1-\frac{1}{n}$ Then $X_{n}$ converges in probability to $0$, so $X_{n}$ converges in distribution to $0$, which implies that $F_{n}$ converges weakly to $F_{0}$, since $F_{n}$ converges completely to $F_{0}$, where $F_{0}$ is the distribution function of a random variable degenerate at $0$.
Now, let $g(x) = x , -\infty < x < \infty$.
Then $\int_{-\infty}^{\infty} g dF_{n} = E(X_{n}) = 0$ and $\int_{-\infty}^{\infty} g dF_{0} = E(0) = 0$.
However, g is not uniformly integrable.
For $a > 1$ and $n\geq 1$,
$\int_{[|x_{n}| \geq a]} |g(x)| dF_{n}(x) = E(|X_{n}| I_{[|x_{n}| \geq a]}) = 0$ if $n < a$ and $= 1$ if $n \geq a$.
Then for $a > 1$,
$\lim_{a \to \infty}\sup_{n \geq 1}\int_{[|x_{n}| \geq a]} |g(x)| dF_{n}(x) = 1$.
Hence, $g$ is not uniformly integrable.
EDIT: According to NCh, this counterexample is valid.
The counter-example is valid. However, the line
should be replaced by $$\int_{[|x| \geq a]} |g(x)| dF_{n}(x) = E(|X_{n}| I_{[|X_{n}| \geq a]}).$$