Problem in measurable set I have two question related to Measure theory ( Lebesgue measure ) any help and hint will be appreciated :-
Problem 1 :- Suppose $E$ and $F$ are subsets of $R$ . Suppose also that the function $s(x)=5\chi_{E}(x)+2\chi_{F}(x)$ is lebesgue measurable. Prove that under the previous conditions, E and F are Lebesgue measurable sets.
Problem 2:- Let $f$ and $g$ be Lebesgue measurable functions on $R$. Prove that the set  $E=\{x\in R :f(x)^2 \ge g(x)^2\}$ is a Lebesgue measurable set.
 A: For the first problem you can use that one definition of measurable function is that for every real number $\alpha$, the set
$$
\{x\in \mathbb{R}: f(x)>\alpha\}
$$
belongs to the $\sigma$-algebra.
For the second, note that the sum and product of two measurable functions is again measurable and that
$$
\{x\in\mathbb{R}:f(x)^2\geq :g(x)^2\}=\{x\in\mathbb{R}:(f(x)-g(x))(f(x)+g(x))\geq 0\}
$$
Edit:
again, for the first problem, note that $E=(E\cap F) \cup (E\setminus F)$ and that $F=(F\cap E)\cup (F\setminus E)$ and that the sets on the right are disjoint. By properties of indicator functions we have 
$$
s(x)= 5\chi_{E\setminus F}(x)+ 7\chi_{E\cap G}+2\chi_{F\setminus E}
$$
If you take $\alpha= 6$ in the definition, the
$$
\{x\in \mathbb{R}: s(x)>6\}= E\cap F
$$
Now, 
$$
E\setminus F=
\{x\in \mathbb{R}: 6\geq s(x)>3\} = 
\{x\in \mathbb{R}: s(x)>3\}\setminus 
\{x\in \mathbb{R}: s(x)>6\}
$$
And the set in the right-hand side of the equation belongs to the $\sigma$-algebra by its properties.
Then you have that $E=(E\cap F) \cup (E\setminus F)$ is measurable. Similarly with $F$
