What are some general strategies to use to get around the issue of non-uniform integrability? Difficulty with exercises in probability theory $\newcommand{\pr}{\operatorname{Pr}}\newcommand{\d}{\,\mathrm{d}}\newcommand{\E}{\mathbb{E}}$I found two nice probability exercises online which at first seemed pretty elementary but I am having a surprising struggle with them:

Let $(\Omega;\pr)$ be a probability space. We are given an almost surely vanishing sequence of nonnegative real random variables $(X_n)_{n\in\Bbb N}$ on $\Omega$ satisfying: $$\forall n\in\Bbb N,\,\E(X_n)=n$$We must prove that $\lim_{n\to\infty}\mathrm{Var}(X_n/n)=\infty$.

And:

Let $(\Omega;\pr)$ be a probability space and $p$ a real number greater than two. Given two sequences of random variables $(X_n)_{n\in\Bbb N},(Y_n)_{n\in\Bbb N}$ with both almost surely vanishing and: $$\forall n\in\Bbb N,\,\E(|X_n|^p+|Y_n|^p)\le1$$We must show that $\E(|X_nY_n|)\to0$ as $n\to\infty$.

I am asking about both because my difficulty with both is essentially the same. I present partial solutions:
To the first:

For any $N\in\Bbb N$ and $\epsilon>0$ define: $$A_{N,\epsilon}:=\bigcap_{m\ge N}\{X_m\le\epsilon\}$$Almost sure convergence to zero implies $\lim_{N\to\infty}\pr(A_{N,\epsilon})=1$ for all $\epsilon$. Moreover, for any $n\ge N$ we know $\{X_n>\epsilon\}\subseteq\Omega\setminus A_{N,\epsilon}$. Then we can bound, for any fixed $n\ge N$, $\epsilon$ and $\frac{\epsilon}{n}<\alpha<\beta$: $$\begin{align}n&=\int_{X_n\le\epsilon}X_n\d\pr+\int_{\epsilon<X_n<n\alpha}X_n\d\pr+\int_{n\alpha\le X_n\le n\beta}X_n\d\pr+\int_{X_n>n\beta}X_n\d\pr\\&\le\epsilon+n\alpha\cdot\pr(\Omega\setminus A_{N,\epsilon})+n\beta\cdot\pr(X_n\ge n\alpha)+\int_{X_n>n\beta}X_n\d\pr\end{align}$$Hence: $$\E\left(\left(\frac{1}{n}X_n\right)^2\right)\ge\alpha^2\cdot\pr(X_n\ge n\alpha)\ge\frac{\alpha^2}{\beta}-\frac{\epsilon\alpha^2}{n\beta}-\frac{\alpha^3}{n\beta}\cdot\pr(\Omega\setminus A_{N,\epsilon})-\frac{\alpha^2}{n\beta}\int_{X_n>n\beta}X_n\d\pr$$So if we can make the RHS as large as desired, for some $N$ and all $n\ge N$ and suitable choices of $\alpha,\beta,\epsilon$, then we are done.

So, my only real idea was that, since $\E(X_n)=n$ coupled with the fact that $X_n\to0$ almost surely should imply that $X_n$ should be large with a relatively high probability (otherwise the expectation would be made too small) then I surely have to show $\alpha^2\cdot\pr(X_n\ge n\alpha)$ can be made as large as desired. This was my focus for a while. I found it difficult to find lower bounds of that quantity since I am not well versed in 'tricks' that go beyond simple manipulations + the use of the Tschebyshev inequality. The above was my best attempt. However, it present a serious issue:
For any $n$ there will be suitably large $\beta_n$ which can make $\int_{X_n>n\beta_n}X_n\d\pr$ small. BUT, since the sequence $(X_n)_{n\in\Bbb N}$ is not uniformly integrable, I cannot fix a choice of $\beta_n$. I'll skip the messy inequality chasing, but I can say with confidence that in the above I can make every term as small as desired whilst keeping $\alpha^2/\beta$ large... except for $\frac{\alpha^3}{\beta}\pr(\Omega\setminus A_{N,\epsilon})$. The trouble is, $\epsilon$ and $N$ have to be fixed first (to make $\pr(\Omega\setminus A_{N,\epsilon})$ suitably small) but for general $n\ge N$ the values $\beta_n$ might be very very large and hence, if I want $\alpha^2/\beta$ to be sufficiently large, the ratio $\alpha^3/\beta$ could blow up.
I would like help learning how to get a "better control" on the size of things like $\int_{X_n>n\beta}X_n\d\pr$ when as $n$ varies, essentially. On one hand, I'd like $\int X_n$ to be made small, but on the other hand the whole point of this problem is that $\int X_n$ is really quite big - I have failed to balance things.
For the other one, the issue is similar:

We can show that if $\epsilon>0$ and $\min(|a|,|b|)>\epsilon$ then $|ab|<\epsilon^{2-p}(|a|^p+|b|^p)$ when $p>2$. Put $Z_n:=|X_n|^p+|Y_n|^p$ for every $n$. We could start (for all $\epsilon>0$): $$\E(|X_nY_n|)\le\epsilon^2+\epsilon^{2-p}\int_{\min(|X_n|,|Y_n|)>\epsilon}Z_n\d\pr$$And the condition that both sequences vanish almost surely ensures $\pr(\min(|X_n|,|Y_n|)<\epsilon)$ can be made as small as desired for all large $n\ge N$. We also know $\E(Z_n)<1$, so we might hope that this gives a way to choose clever values of $\epsilon$ and $N$ (noting that $\epsilon^{2-p}\to\infty$ as $\epsilon\to0^+$, so care is required) such that $\epsilon^{2-p}\int Z_n\d\pr$ is controlled to be small. If so, then we would be done.

But again the trouble is, we don't know whether or not the sequence $(Z_n)_{n\in\Bbb N}$ is uniformly integrable. Without this extra assumption, I don't see how to get the necessary control.
 A: $\newcommand{\bP}{\operatorname{Pr}}$
Concerning the first exercise, it seems easier to work with convergence in probability (which follows from almost sure convergence), that is, for any $\delta > 0$,
$\lim_{n\to \infty} \bP (|X_{n}| > \delta) = 0$.
In particular, $\varepsilon_{n} := \bP (|X_{n}| > 1)$ forms a null sequence
with
\begin{equation}
\label{equ:prob_bound}
(\ast)
\qquad
\bP (|X_{n}| \leq 1) = 1 - \varepsilon_{n},
\qquad
n\in\mathbb{N}.
\end{equation}
Now, for each $n$ separately, let us construct random variables $Y_{n}$ with the lowest variance possible that satisfy $(\ast)$ as well as $\mathbb{E}(Y_{n}) = n$.
These are given by the discrete random variables
[$Y_{n} = 1$ with probability $(1-\varepsilon_{n})$ and $Y_{n} = C_{n}$ with probability $\varepsilon_{n}$], where $C_{n}>0$ is a "large value" that ensures $\mathbb{E}(Y_{n}) = n$, namely (by a simple calculation)
$$
C_{n} = (n+\varepsilon_{n}-1)/\varepsilon_{n}.
$$
Again, $Y_{n}$ is the random variable with the lowest variance possible satisfying the above properties - if we "smear it out" from this discrete choice the variance can only grow (I hope you get my meaning; I gave a proof in my update below using the law of total variance).
In particular, $Y_{n}$ has a lower variance than $X_{n}$.
Hence,
\begin{align*}
\mathrm{Var}(X_{n}/n)
&\geq
n^{-2} \mathrm{Var}(Y_{n})
\\
&=
n^{-2} \big( (1-\varepsilon_{n}) (1-n)^2 + \varepsilon_{n} (C_{n}-n)^2 \big)
\\
&\geq
n^{-2} \, \varepsilon_{n} (C_{n}-n)^2.
\end{align*}
Now plug in the formula for $C_{n}$ to see that this term behaves roughly like $\varepsilon_{n}^{-1}$ as $n\to\infty$, which diverges to $+\infty$.
UPDATE: Proof of the intuitive statement that $Y_{n}$ given above is a random variable with minimal variance given the conditions above, which is an application of the law of total variance:
For a fixed $n$, we start with any random variable $X$ satisfying these conditions.
Similar to Oliver Díaz' idea, we consider the events $A = [X \leq 1]$ and $B = [X > 1]$ and define a corresponding random variable $Z$ by $Z(\omega) = 0$ for $\omega \in A$ and $Z(\omega) = 1$ for $\omega \in B$ in order to apply the law of total variance:
$$
\mathrm{Var}(X)
=
\mathbb{E}(\mathrm{Var}(X|Z)) + \mathrm{Var}(\mathbb{E}(X|Z)).
$$
Now we have precisely the decomposition that we want.
The first term is minimized and equals zero if both $\mathrm{Var}(X|A)$ and $\mathrm{Var}(X|B)$ are zero, i.e. if $X$ is constant on $A$ and $B$, respectively (I hope my notation makes sense to you).
The second term does not really care how the distribution of $X$ is ``smeared out'':
Define $E_{A} = \mathbb{E}(X|A) \leq 1$ and $E_{B} = \mathbb{E}(X|B) \geq 1$, which now have to satisfy the condition
$$
n \stackrel{!}{=} \mathbb{E}(X)
=
E_{A} \bP(A) + E_{B} \bP(B)
=
E_{A} \varepsilon_{n} + E_{B} (1-\varepsilon_{n}).
$$
So, $\mathbb{E}(X|Z)$ is a discrete random variable taking the values $E_{A}\leq 1$ and $E_{B}\geq 1$ with probabilities $\varepsilon_{n}$ and $1-\varepsilon_{n}$, respectively, and satisfying this condition. Hence, its variance $\mathrm{Var}(\mathbb{E}(X|Z))$ is minimized if $E_{A}$ and $E_{B}$ are as close as possible, arriving at $E_{A} = 1$ and $E_{B} = C_{n}$.
Putting things together, we arrive at our random variable $Y_{n}$.
A: This is to complement the solution by @iljusch  to part (a) of the OP's problem. For simplicity, we removed the subscripts $n$ from $X_n$, $Y_n$ and $\varepsilon_n$.
Notice that
$$n=E[X]=E[X\mathbb{1}_{(0,1]}(X)]+E[X\mathbb{1}_{(1,\infty)}(X)]\leq 1-\varepsilon+E[X;X>1]$$
Hence
$$ n-1+\varepsilon\leq E[X;X>1]\leq (E[X^2])^{1/2}\varepsilon^{1/2}$$
and so,
$$\frac{(n-1+\varepsilon)^2}{\varepsilon}\leq E[X^2]$$
It follows that
\begin{align}
\operatorname{var}[Y]=E[Y^2]-n^2&=1-\varepsilon+\frac{(n-1+\varepsilon)^2}{\varepsilon}-n^2\\
&\leq 1-\varepsilon + E[X^2]-n^2\\
&= 1-\varepsilon + \operatorname{var}[X]
\end{align}
Putting things together
\begin{align}
\operatorname{var}[X_n/n]&=\frac1{n^2}\operatorname{var}[X_n]\geq\frac{1}{n^2}(\operatorname{var}[Y_n]-1+\varepsilon_n)\\
&=\frac{1}{\varepsilon_n}\big(\frac{n-1+\varepsilon_n}{n}\big)^2-1\xrightarrow{n\rightarrow\infty}\infty
\end{align}

Edit:
Here is another solution:
with $A=\{X\leq 1\}$ and $Z=\mathbb{1}_A$, $\sigma(Z)=\{\Omega,\emptyset,A,A^c\}$
As @iljusch noted
$$\operatorname{var}[X]\geq\operatorname{var}[E[X|Z]]$$
A simple calculation shows that
$$W:=E[X|Z]=a\mathbb{1}_A+b\mathbb{1}_{A^c}$$
where $a=\frac{E[X\mathbb{1}_A]}{1-\varepsilon}\leq1$ and
$b=\frac{E[X\mathbb{1}_{A^c}]}{\varepsilon}=\frac{n-E[X\mathbb{1}_A]}{\varepsilon}>1$, and
$$\operatorname{var}(E[X|Z])=a^2(1-\varepsilon)+b^2\varepsilon-n^2=
\frac{(E[X\mathbb{1}_A])^2}{1-\varepsilon} + \frac{(n-E[X\mathbb{1}_A])^2}{\varepsilon}-n^2
$$
Hence
\begin{align}
\operatorname{var}[X_n/n]\geq\frac{1}{n^2}\operatorname{var}[E[X_n|Z_n]]\geq \frac{1}{n^2}\frac{(E[X_n\mathbb{1}_{\{Z_n=1\}}])^2}{1-\varepsilon_n}+\frac{1}{n^2}\frac{(n-1)^2}{\varepsilon_n}-1\xrightarrow{n\rightarrow\infty}\infty
\end{align}
since
$$\frac{1}{n^2}\frac{(E[X_n\mathbb{1}_{\{Z_n=1\}}])^2}{1-\varepsilon_n}\leq\frac{1-\varepsilon_n}{n^2}\xrightarrow{n\rightarrow\infty}0$$
and
$$
\frac{1}{n^2}\frac{(n-1)^2}{\varepsilon_n}\xrightarrow{n\rightarrow\infty}\infty
$$
Remark:
Observe that $E[W]=E\big[E[X|Z]\big]=n$ and $P[W\leq1]=P[A]=1-\varepsilon$ and so $W$ satisfies the mean and tail properties that $X$ does.
Notice that $0<t=E[X\mathbb{1}_A]\leq 1-\varepsilon$ and that
$$\phi(t)=\frac{t^2}{1-\varepsilon}+\frac{(n-t)^2}{\varepsilon}$$
is monotone decreasing on $[0,1-\varepsilon]$. Hence, by defining
$$Y=\mathbb{1}_A+C_\varepsilon\mathbb{1}_{A^c}$$
where $C=\frac{n-1+\varepsilon}{\varepsilon}$ we have that $Y$ satisfies $E[Y]=n$ and $E[Y\leq1]=P[A]=1-\varepsilon$. Moreover,
$$E[W^2]\geq E[Y^2]$$
which shows the optimality of $Y$, i.e.
\begin{align}
 Y&=\operatorname{arg}\min\operatorname{var}[X]\\
\text{s.t.} &\quad X\in L^+_2,\, E[X]=n,\, P[X>1]=\varepsilon
\end{align}
A: Here is a solution to the second problem in the OP:
The assumption $X_n,Y_n\xrightarrow{n\rightarrow\infty}0$ implies that
$$\lim_nP[|X_n|>1]=0=\lim_nP[|Y_n|>1],$$
and, by  dominated convergence,
\begin{align}
E[|X_n|^2\wedge1]+E[|Y_n|^2\wedge1]\xrightarrow{n\rightarrow\infty}0\tag{1}\label{one}
\end{align}
Omitting the subscript for the moment and write
\begin{align}
E[|XY|]&=E[|X|\mathbb{1}_{|X|\leq1}|Y|\mathbb{1}_{|Y|\leq1}]+E[|X|\mathbb{1}_{|X|\leq1}|Y|\mathbb{1}_{|Y|>1}]\\
&\qquad+
E[|X|\mathbb{1}_{|X|>1}|Y|\mathbb{1}_{|Y|\leq1}]+
E[|X|\mathbb{1}_{|X|>1}|Y|\mathbb{1}_{|Y|>1}]\tag{2}\label{two}
\end{align}
The first term on the right side of \eqref{two} can be estimated as follows:
$$E[|X_n|\mathbb{1}_{|X_n|\leq1}|Y|\mathbb{1}_{|Y_n|\leq1}]\leq \frac{1}{2}\big(E[|X_n|^2\wedge 1]+E[|Y_n|^2\wedge1])\xrightarrow{n\rightarrow\infty}0
$$
For the second term on the right side of \eqref{two} we have:
\begin{align}
E[|X_n|\mathbb{1}_{|X_n|\leq1}|Y|\mathbb{1}_{|Y_n|>1}]&\leq \frac12\big(E[|X_n|^2\wedge 1]+E[|Y_n|^2;|Y_n|>1]\big)\\
&\leq\frac12\big(E[|X_n|^2\wedge 1]+E[|Y_n|^p]^{2/p}(P[|Y_n|>1])^{1-\tfrac2p}\big)\\
&\leq\frac12\big(E[|X_n|^2\wedge 1]+(P[|Y_n|>1])^{1-\tfrac2p}\big)\xrightarrow{n\rightarrow\infty}0
\end{align}
since $p>2$, $E[Y^p_n]\leq1$, and $Y_n\xrightarrow{n\rightarrow\infty}0$ $p$-a.s.
A similar reasoning shows that the third term on the right side go \eqref{two} converges to $0$ as $n\rightarrow\infty$.
Finally,  for the fourth term on the right side of \rqref{two} we have:
\begin{align}
E[|X_n|\mathbb{1}_{|X_n|>1}|Y|\mathbb{1}_{|Y_n|>1}]&\leq\frac12\big(E[|X_n|^2;|X_n|>1]+E[|Y_n|^2;|Y_n|>1]\big)\\
&\leq\frac12\big(E[|X_n|^p]^{2/p}(P[|X_n|>1])^{1-\tfrac2p}\big) + E[|Y_n|^p]^{2/p}(P[|Y_n|>1])^{1-\tfrac2p}\big)\\
&\leq\frac12\big((P[|X_n|>1])^{1-\tfrac2p} + (P[|Y_n|>1])^{1-\tfrac2p}\big)\xrightarrow{n\rightarrow\infty}0
\end{align}
