Convergence of Types Theorem 
(Convergence of Types Theorem) Suppose that $F_n(u_nx+v_n) \Rightarrow F(x)$ and $F_n(a_nx+b_n) \Rightarrow G(x)$, where $u_n>0, a_n>0$ and $F$ an $G$ are non-degenerate. Then there exist $a>0$ an $b \in\mathbb R$ such that $a_n/u_n \to a$ and $(b_n-v_n)/u_n \to b$, and $F(ax+b) = G(x).$

In Billingsley's textbook, he proves the above theorem using the following lemmas.

  
*
  
*If $F_n \Rightarrow F$, $a_n \to a$ and $b_n \to b$, then $F_n(a_nx +b_n) \Rightarrow F(ax+b)$.
  
*If $F_n \Rightarrow F$ and $a_n \to \infty$, then $F_n(a_nx) \Rightarrow \Delta(x)$, where $\Delta(x)$ is the degenerate distribution at $x$.
  
*If $F_n \Rightarrow F$ and $b_n$ is unbounded, then $F_n(x+b_n)$ cannot converge weakly.
  
*If $F_n(x) \Rightarrow F(x)$ and $F_n(a_nx+b_n) \Rightarrow G(x)$, where $F$ and $G$ are non-degenerate, then $$ 0<\inf_n a_n \leq \sup_n a_n < \infty;\; \sup_n |b_n| < \infty .$$
  

I have difficulty in understanding the proof of the forth lemma (highlighted parts). The argument in the book is as follows. Suppose that $a_n$ is not bounded above. Arrange by passing to a sub-sequence that $a_n \to \infty$. Then by lemma 2, 
$$F_n(a_nx) \Rightarrow \Delta(x).(*)$$
Then since 
$$F_n\left[a_n \left(x+\frac{b_n}{a_n}\right)\right] = F_n(a_nx+b_n)\Rightarrow G(x),(**)$$
it follows by lemma 3 that $\frac{b_n}{a_n}$ is bounded. [Note that in lemma 3, we did not have $a_n$ in front of $(x+b_n)$. But we do now. My question is how to use lemma 3 to get the desired boundedness of $b_n/a_n$, please?] By passing to a further sub-sequence, arrange that $b_n/a_n$ converges to some $c$. By $(*)$ and lemma 1,$F_n\left[a_n \left(x+\frac{b_n}{a_n}\right)\right] \Rightarrow \Delta(x+c)$ along this sub-sequence. But $(**)$ now implies that $G$ is degenerate, contrary to hypothesis. Thus $a_n$ is bounded above. If $G_n(x) = F_n(a_nx+b_n)$, then $G_n(x) \Rightarrow G(x)$ and $G_n(a_n^{-1}x - a_n^{-1}b_n) = F_n(x) \Rightarrow F(x)$. The result just proved shows that $a_n^{-1}$ is bounded. Thus $a_n$ is bounded away from $0$ and $\infty$. My question here is how to know $a_n$ is positive, please? How come $a_n$ cannot be negative? Thank you!

It seems that Lemma 3 is applied with $\widetilde{F_n}(t):=F_n(a_nt)$ instead of $F_n$ (which is allowed by Lemma 2) and the sequence $(b_n/a_n)_{n\geqslant 1}$ instead of $(b_n)_{n\geqslant 1}$.  
