Finding the right conditions for a sequence to satisfy convergence in distribution then convergence in probability then convergence almost surely I couldn't solve this exercice so I could use some help. It goes like this : 
Let $(X_n)_{n \in\Bbb N^*}$ be a sequence of real independent random variable defined in some probability space. Let $F_{X_n}$ be the cumulutive distribution function such that
$$F_{X_n}=\begin{cases}
1 & \quad   x > 1\\
a_n+(1-a_n)x^n & \quad x \in[0,1] \\ 
0 & \quad  x <0
\end{cases}$$
$a_n\in [0,1] $
Question :
Find the right conditions for $a_n$ to have :
1/ Convergence in distribution.
2/ Convergence in probability.
3/Almost sure convergence.
( note that the questions are seperated so it needs to be solved one at a time ).
 A: 1/ Convergence in distribution: We have $X_n$ converges in distribution iff $F_{X_n}(x)$ converges for all $x$.
For $x \in [0,1)$, as $a_n \in [0,1]$ then $(1-a_n)x^n \xrightarrow{n \to +\infty} 0 $, so
$F(x)$ converges for $x \in [0,1)$ iff the sequence $a_n$ converges to $a$. And because $F_{X_n}(x) \in [0,1]$ then $a \in [0,1]$.
We check this condition $a \in [0,1]$. Indeed, with this condition, $X_n$ converges to the distribution $$X with ( $P(X=0) = a$ and $P(X=1) = 1-a$)
Hence, $X_n$ converges in distribution if and only if the sequence $a_n$ converges to $a \in [0,1]$.
2/ Convergence in probability:
As the convergence in probability implies the convergence in distribution, then a necessary condition of the convergence in probability is the sequence $a_n$ converges to $a \in [0,1]$. And so, $X_n$ converges to $X$ with ($P(X=0) = a$ and $P(X=1) = 1-a$).
It suffices to verify if this condition is sufficient. We need
$$P(|X_n-X| \ge \epsilon) \xrightarrow{n \to +\infty} 0 $$
We have
\begin{align}
P(|X_n-X| \ge \epsilon) & = P(\{|X_n-X| \ge \epsilon \} \cap \{ X = 0\}) +P(\{|X_n-X| \ge \epsilon \} \cap \{ X = 1\}) \\
&= P(X=0)P(X_n \ge \epsilon ) +P(X=1)P( 1-X_n \ge \epsilon ) \\
&= a(1-F_{X_n}(\epsilon)) +(1-a)F_{X_n}(1-\epsilon) \\
&= a(1-a_n-(1-a_n)\epsilon^n) +(1-a)(a_n + (1-a_n)(1-\epsilon)^n) \tag{1}
\end{align}
Then
$$P(|X_n-X| \ge \epsilon) \xrightarrow{n \to +\infty} a(1-a) + (1-a)a = 2a(1-a)$$
By consequence, we can conclude that the $X_n$ converges in probability if and only if the sequence $a_n$ converges to $a = 0$ or $a=1$. And $X_n$ converges to the constant $X = 1-a$.
3/ Convergence almost sure:
As the convergence almost sure implies the convergence in probability then a necessary condition for the almost sure convergence is the sequence $a_n$ converges to $a = 0$ or $a = 1$ .
We need to check if this condition is sufficient by using the Borel-Cantelli Lemma. Let's denote
$$A_n(\epsilon) = \{ \omega: |X_n(\omega) - 0| > \epsilon\} $$
From $(1)$, we have

*

*If $a = 0$: $$P(A_n(\epsilon)) = a_n + (1-a_n)(1-\epsilon)^n$$ $$\implies \sum_{n=1}^N P(A_n(\epsilon)) =\sum_{n=1}^N a_n + \sum_{n=1}^N (1-a_n)(1-\epsilon)^n $$ We notice that if the sequence $a_n$ converges then $\sum_{n=1}^N (1-a_n)(1-\epsilon)^n < +\infty$, then $\sum_{n=1}^N P(A_n(\epsilon)) <+\infty$ if and only if $$\sum_{n=1}^N a_n<+\infty \tag{2}$$

*If $a = 1$: $$P(A_n(\epsilon)) = (1-a_n)(1-\epsilon^n)$$ $$ \implies \sum_{n=1}^N P(A_n(\epsilon)) = \sum_{n=1}^N (1-a_n)+  \sum_{n=1}^N (1-a_n)\epsilon^n $$ As $X_n$ independent, by applying the Borel-Cantelli Lemma (part b), if the sum $\sum_{n=1}^{+\infty} (1-a_n) = \infty$ then $X_n$ can't converge to $X = 0$. Hence, $\sum_{n=1}^N P(A_n(\epsilon)) <+\infty$ if and only if
$$\sum_{n=1}^{+\infty} (1-a_n) < +\infty \tag{3}$$  (it's evident that $\sum_{n=1}^N (1-a_n)\epsilon^n <+\infty$)

And by applying the Borel-Cantelli Lemma (part a), we have $(2)$ and $(3)$ are also the sufficient condition for the almost sure convergence.
By consequence, $X_n$ converges almost surely if and only if one of these conditions occurs $$\sum_{n=1}^{+\infty} (1-a_n) < +\infty$$
$$\sum_{n=1}^{+\infty} a_n < +\infty$$
A: 1/ Suppose that $X_n$ converges in distribution to $X$. Then $F_{X_n}$ converges almost everywhere to $F_X$. In particular there exist $x,y\in(0,1)$, $x\neq y$ such that the sequences $x_n=a_n+(1-a_n)x$ and $y_n=a_n+(1-a_n)y$ converge. Since $a_n=(yx_n-xy_n)/(y-x)$ we deduce that $a_n$ must converge to some $a\in\mathbb R$. Since $a_n\in[0,1]$ we even have $a\in[0,1]$.
Conversely suppose that $a_n$ converges to some $a\in[0,1]$. Then $F_{X_n}$ converges everywhere to the map $F$ defined by $F(x)=0$ if $x<0$, $F(x)=1$ if $x>1$ and $F(x)=a+(1-a)x^n$ if $x\in[0,1]$. It is clear that $F$ is nondecreasing from $0$ to $1$ and right-continuous. Therefore there exists a random variable $X$ whose $F$ is the distribution function (take $X=F^{-1}(U)$ where $F^{-1}$ is the left pseudo-inverse of $F$ and $U$ uniformly distributed on $(0,1)$). Since $F_{X_n}$ converges on $\mathbb R$ to $F_X$ we deduce that $X_n$ converges in distribution to $X$.
We deduce that there is convergence is distribution iff $a_n$ is convergent.
3/ Suppose that $X_n$ converges almost surely to $X$. According to 1/, we saw that $a_n$ converges to $a\in[0,1]$ and $F_{X_n}$ converges to $F_X$ almost everywhere. Since $X_n$, $n\in\mathbb N^*$ are independent, it is well known that the almost sure limit $X$ is almost surely constant, see for instance Independence is preserved under taking almost sure limits applied with $Y_n=X_{n+1}$. So $X$ is almost surely constant. In other words, there exists $y\in[0,1]$ such that $F_X(y-)=0$ and $F_X(y)=1$. Since $F_X(y)=a+(1-a)x^n$ for all $x\in[0,1]$, this can only happen if $a=1$, in which case $X$ is almost surely constant equal to $0$.
Conversely, suppose that $a_n$ converges to $1$. Let $\varepsilon\in(0,1)$. For all $n\in\mathbb N^*$ we have $\mathbb P(X_n>\varepsilon)=1-F_{X_n}(\varepsilon)=1-a_n-(1-a_n)\varepsilon^n=(1-a_n)(1-\varepsilon^n)$. Therefore $\sum_{n\in\mathbb N^*}\mathbb P(X_n>\varepsilon)$ converges iff $\sum_{n\in\mathbb N^*}(1-a_n)$ converges.
If $\sum_{n\in\mathbb N^*}(1-a_n)$ converges, then by the Borel-Cantelli lemma, $\mathbb P(\bigcap_{N\in\mathbb N^*}\bigcup_{n\ge N}\{X_n>\varepsilon\})=0$. Equivalently, we have almost surely $\bigcup_{N\in\mathbb N^*}\bigcap_{n\ge N}\{X_n\le\varepsilon\}$. By taking for instance $\varepsilon=1/p$, this shows that almost surely, for all $p\in\mathbb N^*$, there exists $N\in\mathbb N^*$ such that for all $n\ge N$, $X_n\le1/p$, hence $X_n$ converges almost surely to $0$.
If $\sum_{n\in\mathbb N^*}(1-a_n)$ does not converge, then by the Borel-Cantelli lemma again, $\mathbb P(\bigcap_{N\in\mathbb N^*}\bigcup_{n\ge N}\{X_n>\varepsilon\})=1$. We deduce that almost surely, for all $N\in\mathbb N^*$ there exists $n\ge N$ such that $X_n>\varepsilon$, which contradicts that $X_n$ converges almost surely to $0$.
We deduce that there is almost sure convergence iff $\sum_{n\in\mathbb N^*}(1-a_n)$ converges.
2/ Suppose that $X_n$ converges in probability. It is well known that convergence in probability implies convergence in distribution, and also almost sure convergence of a subsequence. From 1/ we deduce that $a_n$ converges to some $a\in[0,1]$, and from 2/ we deduce that there exists a subsequence of $(a_n)_{n\in\mathbb N^*}$ which converges to $1$, hence $a=1$ and $a_n$ converges to $1$.
Conversely, suppose that $a_n$ converges to $1$. From 1/ we deduce that $X_n$ converges in distribution to a variable which is almost surely constant to $0$. It is well known that convergence in distribution implies convergence in probability as soon as the limit is almost surely constant. Hence $X_n$ converges in probability.
We deduce that there is convergence in probability iff $a_n$ converges to $1$.
