# Convergence in the absence of Dominated Convergence Theorem, and uniform integrability

This question is extended from Resnick's exercise 5.13 in his book A Probability Path.

Let the probability space be the Lebesgue interval, that is, $(\Omega=[0,1],\mathcal{B}([0,1]),\lambda)$ 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)

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I) $X_{n} \rightarrow 0$. You should use that $X_{n}(w) \neq 0$ iif $w \in (0,1/n)$. The proof follows almost from definition.

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 $\lim X_{n}(w) = 0$.

II) $E[X_{n}] \rightarrow 0$. Observe that $X_{n}$ is constant and therefore, it is easy to compute $E[X_{n}]$.

$E[X_{n}] = \frac{n}{\log(n)}P(Y \in (0,1/n)) = \frac{1}{\log(n)}$.

III) Uniform integrability. Observe that $X_{n} \leq n$. 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.

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