This question is extended from Resnick's exercise 5.13 in his book A Probability Path.
Let the probability space be the Lebesgue interval:
$(\Omega=[0,1],\mathcal{B}([0,1]),\lambda)$ and define
$X_n:=\frac{n}{\log n}1_{(0,\frac 1n)}$
Show $X_n\to 0, E(X_n)\to 0$ even though DCT fails. And secondly, show
$\lim_{M\to\infty} \sup_{n\ge 2} E(X_n 1_{X_n>M})=0$ (uniform integrability)
Attempt/outline at a solution
:$X_{n} \rightarrow 0$. You should use that $X_{n}(w) \neq 0$ iif $w \in (0,1/n)$.
:For any $w \in (0,1)$, there exists $n^{*}$ such that $\frac{1}{n^{*}} < w$. Hence, for any $n > n^{*}$, $X_{n}(w) = 0$ and $\lim X_{n}(w) = 0$.
:$E[X_{n}] \rightarrow 0$. $X_{n}$ is constant and therefore,
:$E[X_{n}] = \frac{n}{\log(n)}P(Y \in (0,1/n)) = \frac{1}{\log(n)}$.
:Uniform integrability. $X_{n} \leq n$. Hence,
$\sup_{n \geq 2} E[X_{n}I(X_{n} > M)] \geq \sup_{n \geq M} E[X_{n}I(X_{n} > M)] \geq \sup_{n \geq M} E[X_{n}] = \sup_{n \geq M}\frac{1}{\log(n)} = \frac{1}{\log(M)}$
The result follows taking the limit in M.
Critique of the solution
Implicit in Resnick's instructions is that you explain why you cannot use DCT. (Hint: think about $Y = \sup_{n\ge2}X_n$ and its expectation.)
Your check of uniform integrability is wrong. To begin with, you have to bound the expression from above not from below. Second, since you have a supremum (over all n), the expression can only go to 0 because of the indicator for X_n>M; your argument appears to take no notice of it. Keep in mind that X_n is positive only on (0, 1/n) so no X_n dominates another.