Exam Problem from Probability Class

These two days PG students in our office struggled for their take-home final exam with dead-line two days after it was released. They were stuck with the following problem (and so I was, pleasantly, asked for help):

Problem. Let $(X,\mu)$ be a probability measure space such that $f_n \stackrel{\mu}{\to} 0$ and that $\text{Var}(f_n)=1$, prove that $E(f_n)\to 0$.

Here $\text{Var}(f_n) = \int_X f_n^2\,d\mu -(\int_X f_n\,d\mu)^2$ and $E(f_n)=\int_X f_n\,d\mu$.

In the sequel we denote $\int f=\int_Xf\,d\mu$ and $\int_A f=\int_A f\,d\mu$. Then $E(f_n)$ is also written as $\int f_n$.

In the proof below we need an elementary inequality from probability that easily follows from Chebyshev's inequality: $$\mu\{x\in X:|f_n(x)-E(f_n)|\ge a\}\leq \frac{\text{Var}(f_n)}{a^2}.$$
Proof.
Suppose on the contrary that $E(f_n)\not\to 0$, then there is an $\epsilon_0>0$ such that $|\int f_{n_k}|\ge \epsilon_0$. It follows that $$\begin{align*} \epsilon_0&\leq \left|\int_{\{|f_{n_k}|\ge 1\}} f_{n_k}\right|+\left|\int_{\{|f_{n_k}|< 1\}} f_{n_k}\right|\\ &\leq\sqrt{\int_{\{|f_{n_k}|\ge 1\}}  f_{n_k}^2} \cdot \sqrt{\mu\{|f_{n_k}|\ge 1\} } + \int|f_{n_k}|\chi_{\{|f_{n_k}|<1\}}\\ &\leq \sqrt{\int f_{n_k}^2}\cdot  \sqrt{\mu\{|f_{n_k}|\ge 1\} } + \int|f_{n_k}|\chi_{\{|f_{n_k}|<1\}}. \end{align*}$$ Since $f_{n_k}\to 0$ in measure, there is a subsequence $f_{n_{k_p}}$ that converges to 0 pointwise $\mu$-a.e.. By Lebesgue Dominated Convergence Theorem, $\int|f_{n_{k_p}}|\chi_{\{|f_{n_{k_p}}|<1\}}\to 0$. So the above becomes $$\epsilon_0\leq \lims_{p\to \infty} \sqrt{\int f_{n_{k_p}}^2}\cdot  \sqrt{\mu\{|f_{n_{k_p}}|\ge 1\} } .$$ We will finish the proof by showing $\int f_n$ is in fact bounded in $n$, and so is $\int f_n^2$ since $1=\text{Var}(f_n)=\int f_n^2 - (\int f_n)^2$, and then the above inequality yields a contradiction that $\epsilon_0\leq 0$.

Now we turn to the boundedness of  $\{\int f_n\}$, the only tricky part of this problem. For this, observe that for any measurable $f,g$ on $X$, $$\mu\{|f|\ge 2\epsilon\}\leq \mu \{|f-g|+|g|\ge 2\epsilon\}\leq \mu\{|f-g|\ge \epsilon\}+\mu\{|g|\ge \epsilon\}.$$ In particular, let's take $f$ to be the constant function $E(f_n)$, and $g=f_n$, then $$\mu\{x\in X: |E(f_n)|\ge 2\epsilon\} \leq \mu \{|E(f_n)-f_n|\ge \epsilon\} + \mu\{|f_n|\ge \epsilon\} \leq \frac{1}{\epsilon^2} + \mu\{|f_n|\ge \epsilon\},$$ the last inequality follows from $\mu \{|f_n-E(f_n)|\ge \epsilon\} \leq \frac{\text{Var}(f_n)}{\epsilon^2}$. Taking $\lims$ on both sides above, we get $$\begin{equation}\label{gen it} \lims \mu\{x\in X: |E(f_n)|\ge 2\epsilon\} \leq \frac{1}{\epsilon^2} \end{equation}$$ by convergence in measure. Take $\epsilon=2$, there is an $N$ such that for every $n>N$, $$\mu\{x\in X: |E(f_n)|\ge 2\cdot 2\}  \bm{ <1},$$ and therefore $\{x\in X: |E(f_n)|\ge 4\}=\emptyset$, and thus $|E(f_n)|< 4$ for $n>N$.$\qed$

Remark. After examining the proof the condition that $\text{Var}(f_n)=1$ can be replaced by that the sequence $\{\text{Var}(f_n)\}_{n=1}^\infty$ is bounded, since we can generalize ($\ref{gen it}$) to
$$\boxed{\dis \lims \mu\{x\in X: |E(f_n)|\ge 2\epsilon\} \leq \frac{\text{Var}(f_n)}{\epsilon^2}.}$$