$\limsup$ of Images of collection of measurable sets Let $\{A_j\}_{j=1}^{\infty}$ be a collection of Lebesgue measurable sets and $\sum\limits_{j=1}^{\infty}\lambda(A_j)<\infty, f(x)=x^2$ then $\lambda(\limsup f(A_j))=0$. $\lambda$ is the lebesgue measure on $\mathbb{R}$.
By definition, $\limsup f(A_j)=\{y \mid y \in f(A_j) \ \text{for  infinitely  many} \ j\}$. Since each $A_j$ is lebesgue measurable, they are of the form $A_j=E_j \cup N_j$ where $E_j$ is borel, and $N_j$ is null. Since $\sum\limits_{j=1}^{\infty}\lambda(A_j)<\infty$, we may assume that the nonull parts of $A_j$ are bounded (Here I mean if one of the $A_i$'s is unbounded, the part that is unbounded has measure zero). Let $y \in f(A_j)$ for some $j$. Since $|f^{-1}(\{y\})|\leq 2$ for each $y$, and $\{A_j\}$ is infinite, we must have each $f^{-1}(\{y\})$ are contained in only finitely many sets of positive measure. Since this holds for any $y \in \bigcup\limits_{j=1}^{\infty}f(A_j)$, we see that any point in the set is contained in only finitely many sets of positive measure so that $\lambda(\limsup f(A_j))=0$.
I have tried this problem several times and have been unsuccessful each time. Is this the correct way to do this? Also, my "proof" does not seem rigorous enough. How to make this more rigorous?
Edit: I see there is a major flaw in this attempt. I cannot assume anything about the boundedness of the $A_i$'s. How should I take another approach to this?
 A: Let $y \in \lim \sup f(A_n)$. Then we can write $y=x_n^{2}$ with $x _n \in A_n$ for infinitely many values of $n$. But then $\pm \sqrt y \in A_n$ for infinitely many values of $n$ which implies either $ \sqrt y \in A_n$ for infinitely many values of $n$ or $- \sqrt y \in A_n$ for infinitely many values of $n$. But $\sum \lambda (A_n) <\infty$ implies that $\lim \sup A_n$ has measure $0$. So there is a set $E$ such that $\lambda (E)=0$ and $\pm \sqrt y \in E$ whenever $y \in \lim \sup f(A_n)$. Can you finish the proof now?
A: Put $B_n = A_n \cup (-A_n)$. We have $$y \in \overline{\lim} f(A_j) \Rightarrow \big(\forall N \exists n \ge N: y \in f(A_n) \big) \Rightarrow $$ $$\big( \forall N \exists n \ge N \exists x_n \in A_n: x_n^2 = y\big) \Rightarrow \big(  \forall N \exists n \ge N \exists x_n \in A_n: x_n = \pm \sqrt{y} \big) $$ $$\Rightarrow \big( \forall N \exists n \ge N: (\sqrt{y} \in A_n) \vee (\sqrt{y} \in - A_n)\big)$$ $$\Rightarrow \big( \forall N \exists n \ge N: \sqrt{y} \in B_n \big) \Rightarrow  \sqrt{y} \in \overline{\lim}  B_n.$$
But $\lambda(B_n) =\lambda(A_n \cup (-A_n)) \le  \lambda(A_n) + \lambda(-A_n) = 2 \lambda(A_n)$ and hence $\sum_n \lambda(B_n) < \infty$. Put $C = \overline{\lim}  B_n$. It follows from Borel–Cantelli Lemma for measure spaces that $\lambda(C) = 0$.
So we proved that $y \in  \overline{\lim} f(A_j) \Rightarrow \sqrt{y} \in C$ and $\lambda(C) = 0$. It follows that $\lambda(\overline{\lim} f(A_j)) = 0$.
Addition:
Lemma. If $\lambda(C) = 0$ and $\big( y \in B \Rightarrow \sqrt{y} \in C \big) $ then $\lambda(B) = 0$.
Proof.
Note that $y \in B \Rightarrow  \sqrt{y}  \exists   \Rightarrow B \subset [0,+\infty)$. Consider $B^* = B \cap (a,b)$ for some positive $a, b > 0$. It's sufficient to show that $\lambda(B^*) = 0$. We know that $\sqrt{B^*} \subset C$. In other words, $g(B^*) \subset C$ where $g$ is strictly increasing, $0 < C_1 \le g' \le C_2 < \infty$ and $g'$ is continious. It's sufficient to use lemma 2.
Lemma 2.
If $\lambda(C) = 0$ and $g(B^*) \subset C$ where $0 < C_1 \le g' \le C_2 < \infty$ and $g'$ is continious. Then $\lambda(B^*) = 0$.
Hint to the proof: we may consider sets $D_j$ from definition of outer measure of $C$ and take $g^{-1}(D_j)$.
