What are the necessary and sufficient conditions for $α_i<β_i$ such that independent $\operatorname{Uniform}(α_i,β_i)$ variables $X_i$ What are the necessary and sufficient conditions for $α_i<β_i$ such that the independent $\operatorname{Uniform}(α_i,β_i)$ distributed variables $X_i$:
(1) converge to $0$ in distribution?
(2) converge to $0$ almost surely?
The $\operatorname{Uniform}(α_i,β_i)$ variables $X_i$ are not identically distributed.
For 1 we could use convergence in probability as it converges to a constant, but I don't know if that will help solve this problem. 
For 2, I believe I need to use Borel-Cantelli lemmas.
 A: Apply the definitions.
Convergence in distribution
To converge to $0$ in distribution, we need the CDFs of the $X_i$ (which I will call $F_i$) to converge pointwise to $1$ at all positive real numbers and to $0$ at all negative real numbers.  Those CDFs are specifically  $$F_i(x) = \left\{\eqalign{0 &\quad x \le \alpha_i \\ \frac{x-\alpha_i}{\beta_i-\alpha_i} &\quad \alpha_i \lt x \lt \beta_i \\ 1 &\quad x \gt \beta_i}\right.$$
When $x \gt 0,$ the convergence of the sequence of (nonnegative) real numbers $(F_i(x))$ means that for any $\epsilon\gt 0,$

*

*there is an $N(x,\epsilon)$ for which $i \ge N(x,\epsilon)$ implies $1 - \epsilon \le F_i(x)$ and,


*similarly, because $-x \lt 0,$ there is an $N(-x,\epsilon)$ for which $i \ge N(-x,\epsilon)$ implies $\epsilon \ge F_i(-x).$
Supposing $\epsilon\lt 1$ and $i$ exceeds both of $N(x,\epsilon)$ and $N(-x,\epsilon),$ we deduce that $\alpha_i \le 0 \le \beta_i$ (for otherwise $F(x)$ is either $0$ or $1$ and one of the inequalities (1) or (2) will be violated) and then the formula for $F_i$ gives
$$ 1-\epsilon \le\frac{x-\alpha_i}{\beta_i-\alpha_i}\text{ and } \frac{-x-\alpha_i}{\beta_i-\alpha_i} \le \epsilon.$$
Equivalently, writing $\delta_i = \beta_i - \alpha_i \gt 0$ for convenience,
$$\delta_i(1-\epsilon) - x \le -\alpha_i \le \epsilon \delta_i + x.$$
The inequality between the two outer terms implies
$$\delta_i \le \frac{2x}{1-2\epsilon}.$$
Because $x$ and $\epsilon$ are arbitrarily small positive numbers, it follows that $\delta_i$ must converge to $0.$  It is then easy to see (and show) that both $\alpha_i$ and $\beta_i$ must converge to $0,$ (for otherwise the $X_i$ either do not converge in distribution at all or they converge to some nonzero value).

Convergence of $F_i$ to the distribution of $0$ implies $\alpha_i\to 0$ and $\beta_i\to 0.$

Almost sure convergence
This will occur when $\Pr(\lim_{i\to\infty}X_i = 0) = 1.$  Suppose $\alpha_i\to 0$ and $\beta_i\to 0.$  Together these assumptions imply that for any $\epsilon\gt 0,$ eventually (for large enough $i$) $|\alpha_i| + |\beta_i| \le \epsilon,$ which in turn means $$\Pr(-\epsilon \le X_i \le \epsilon) = 1.$$ That demonstrates almost sure (a.s.) convergence.

$\alpha_i\to 0$ and $\beta_i\to 0$ imply a.s. convergence of $X_i.$

The conclusion
Finally, it is well known that a.s. convergence implies convergence in distribution.  We therefore have achieved the cycle of implications

a.s. convergence implies convergence in distribution, which necessitates (separate) convergence of both $\alpha_i$ and $\beta_i,$ which implies a.s. convergence.

Thus, all three conditions are necessary and sufficient.
A: It's been more than a year since I took my course in graduate probability theory, so this material is not very fresh in my mind, so these answers may not be correct.
Hopefully, they serve as either (1) suggestions which ultimately motivate you to the correct answer or (2) something which prompts someone else to post the correct answer.
2. For convergence almost surely, we want the sequence of random variables $(X_n) \sim \operatorname{Uniform}(\alpha_i,\beta_i)$ to satisfy, for $X \equiv 0$ (constant random variable), $$\mathbb{P}(\{\omega \in \Omega: |X_i(\omega) - X(\omega)| \underset{i \to \infty}{\to} 0\})=1 \,.$$ Since $X\equiv 0$ is a constant random variable, we have that $X(\omega)=0$ for all $\omega \in \Omega$, so we can rewrite the above as $$\mathbb{P}(\{\omega \in \Omega: |X_i(\omega)| \underset{i \to \infty}{\to} 0\})=1 \,.$$ Let's go back to real analysis for a moment and remember what $|X_i(\omega)|\underset{i \to \infty}{\to} 0$ means: for every $\varepsilon >0$, there exists an $I_{\varepsilon} \in \mathbb{N}$ such that for all $i \ge I_{\varepsilon}$, $|X_i(\omega)| < \varepsilon$.
Since $X_i \sim \operatorname{Uniform}(\alpha_i, \beta_i)$, we have that $X_i(\omega) \in (\alpha_i,\beta_i)$ for all $\omega \in \Omega$.
Thus it is a sufficient condition (I'm not sure if it's actually necessary) for almost sure convergence that, for all $\varepsilon > 0$, there exists an $I_{\varepsilon} \in \mathbb{N}$ such that, for all $i \ge I_{\varepsilon}$, $x \in (\alpha_i, \beta_i) \implies |x| < \varepsilon$.
In particular, it is sufficient (again I'm not sure if it's necessary) that $\alpha_i \underset{i \to \infty}{\to} 0$, $\beta_i \underset{i \to \infty}{\to} 0$ (if we want $|x| < \varepsilon$, we can just take the maximum of the indices such that $\alpha_i < \varepsilon/2$, $\beta_i < \varepsilon/2$).
You are probably right that applying Borel-Cantelli would be helpful here; I am sufficiently out of practice that I have lost some of my intuition for how to write a Borel-Cantelli argument, and thus will not attempt to do so here.
1. A necessary condition for convergence in probability is convergence almost surely along some subsequence. 
According to this answer to another question, if every subsequence of the $X_i$ has a further subsequence which converges almost surely to $X$, then the $X_i$ converge in probability to $X$. Thus this is a sufficient condition for convergence in probability.
It might be possible to combine the two above to get a necessary and sufficient condition for convergence in probability from the condition for convergence almost surely, but I'm not certain.
Using an argument analogous to that above for almost sure convergence, it probably would not be difficult to directly apply the definition of convergence in probability to this specific case to get a necessary and sufficient condition.
However, instead let's use your observation that in this case convergence in distribution implies convergence in probability.
The cumulative distribution function of the constant random variable $X \equiv 0$ is continuous everywhere except at $0$ (where it jumps from $0$ to $1$).
Thus, we just need to show that, for all $x \not=0$, the cumulative distribution functions of the $X_i$ converge to the cumulative distribution function of the constant random variable $X \equiv 0$.
If we let $F_i(x)$ denote the cumulative distribution function of $X_i$, then what we need to show is that, for $x < 0$, $F_i(x) \underset{i \to \infty}{\to} 0$, and for $x >0$, $F_i (x) \underset{i \to \infty}{\to } 1$. I leave it to you to formulate this condition in terms of the $\alpha_i$ and the $\beta_i$.
The above condition is necessary and sufficient for convergence in distribution, and since the convergence is to a constant random variable, it is also necessary and sufficient for convergence in probability.
