# Uniform Integrability Counterexample

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.

• Your counterexample is correct. . – NCh Mar 26 '17 at 17:17
• There are no mistakes? I just noticed that I made a mistake when writing out the set of integration for calculating the expected value at the end. I corrected this. – shmiggens Mar 26 '17 at 17:49
• I did not find any mistakes here. – NCh Mar 26 '17 at 17:59

$$\int_{[|x_{n}| \geq a]} |g(x)| dF_{n}(x) = E(|X_{n}| I_{[|x_{n}| \geq a]})$$
should be replaced by $$\int_{[|x| \geq a]} |g(x)| dF_{n}(x) = E(|X_{n}| I_{[|X_{n}| \geq a]}).$$