Probability Qual problem: $X_n\rightarrow X$ in probability implies $EX_n\rightarrow EX$ under given condition?

Question

Suppose $X_n\rightarrow X$ in probability, and there is a continuous function $f$ with $f(x)>0$ for large x with $\dfrac{\vert x\vert}{f(x)}\rightarrow 0$ as $|x|\rightarrow\infty$ so that $Ef(X_n)\leq C<\infty$ for all $n$. Show that $EX_n\rightarrow EX$.

Any hint/proof/approach is welcome, thanks in advance..

Answer 1

Let $\varepsilon>0$. Since $\frac{|x|}{f(x)} \to 0$ as $|x| \to \infty$ we can choose $k=k(\varepsilon)$ such that

$$\frac{|x|}{f(x)} \leq \varepsilon \qquad \text{for all} \, |x| \geq \frac{k}{2} \tag{1}$$

Now fix $m,n \in \mathbb{N}$. Then,

$$\begin{align*} \int |X_n-X_m| \, d\mathbb{P} &= \int_{|X_n-X_m| < \varepsilon} |X_n-X_m| \, d\mathbb{P}+ \int_{|X_n-X_m| \geq \varepsilon, |X_n-X_m| \leq k} |X_n-X_m| \, d\mathbb{P} \\ &\quad +\int_{|X_n-X_m| \geq \varepsilon, |X_n-X_m| > k} |X_n-X_m| \, d\mathbb{P} \\ &\leq \varepsilon + k \cdot \mathbb{P}(|X_n-X_m| \geq \varepsilon) + \int_{|X_n-X_m| > k} |X_n-X_m| \, d\mathbb{P} \tag{2} \end{align*}$$

By the triangle inequality, we find

$$\begin{align*} \int_{|X_n-X_m| > k} |X_n-X_m| \, d\mathbb{P} &\leq 2 \sup_{n \in \mathbb{N}} \int_{|X_n|>k/2} \frac{|X_n|}{f(X_n)} f(X_n) \, d\mathbb{P} \\ &\stackrel{(1)}{\leq} 2\varepsilon \sup_{n \in \mathbb{N}} \int f(X_n) \, d\mathbb{P} \leq 2\varepsilon \cdot C \tag{3} \end{align*}$$

Combining $(2)$ and $(3)$ yields

$$ \int |X_n-X_m| \, d\mathbb{P}\leq \varepsilon + k \cdot \mathbb{P}(|X_n-X_m| \geq \varepsilon) + 2\varepsilon \cdot C \stackrel{n,m \to \infty}{\to} \varepsilon (1+2C) \stackrel{\varepsilon \to 0}{\to} 0$$

using that $\mathbb{P}(|X_n-X_m| \geq \varepsilon) \to 0$ as $m,n \to \infty$ since $X_n \to X$ in probability. Consequently, we have shown that $(X_n)_{n \in \mathbb{N}}$ is a $L^1$-Cauchy sequence. Since $L^1(\mathbb{P})$ is complete, we get $X_n \to Y$ in $L^1$ for some $Y \in L^1$. By the uniqueness of the limit, we conclude $X=Y$ and this finishes the proof.

Answer 2

It suffices to show the $X_n$ are uniformly integrable. Indeed, since $|x|/f(x) \to 0$ we have that $E[|X_n| 1_{|X_n|>\lambda}] \leq k E[f(X_n)] \leq k C$ where $k$ may be chosen to be as small as we like by taking $\lambda$ large enough. It follows that $X_n$ are UI and so Vitali's convergence theorem implies we have $L_1$ convergence.