Let $\chi_{E}(x)$ and $\chi_{F}(x)$ be the characteristic functions of $E$ and $F$, where $E$ and $F$ are measurable sets and $E \cup F = \mathbb{R^n}$. Prove by definition that $f(x) = 6\chi_{E}(x) - 5\chi_{F}(x)$ is measurable.
My solution is:
First we show that $a+b$ is measurable for any measurable functions $a$ and $b$. Since $b$ is measurable, we have the set $\{b(x) \gt \alpha\}$ is measurable $\forall \alpha \in \mathbb{R^n}$. Hence $\{-b(x) \gt \alpha\}$ = $\{b(x) \lt -\alpha\}$ is measurable for all $\alpha$ and so -b is measurable. Since $\{\beta - b(x) \gt \alpha\}$ = $\{b(x) \gt \alpha - \beta\}$ is measurable $\forall\alpha$ since $\alpha - \beta \in \mathbb{R^n}$, we have that $\beta - b(x)$ is measurable for any $\beta$. Thus, since $\{a(x) + b(x) \gt \alpha\}$ = $\{a(x) \gt \alpha - b(x)\}$ which is measurable for any $\alpha$ since $\alpha - b(x)$ is measurable, we have $a+b$ is measurable.
Since $\chi_{E}(x)$ and $\chi_{F}(x)$ are characteristic functions, we know they are measurable. Hence, $\{ \chi_{E}(x) \gt \alpha \}$ and $\{ \chi_{F}(x) \gt \alpha \}$ are measurable $\forall\alpha$. Since $\{ 6\chi_{E}(x) \gt \alpha \}$ = $\{ \chi_{E}(x) \gt \frac{\alpha}{6} \}$ where $\frac{\alpha}{6} \in \mathbb{R^n}$, we have $6\chi_{E}(x)$ is measurable. Similarly, $-5\chi_{E}(x)$ is measurable.
Thus, $f(x) = 6\chi_{E}(x) - 5\chi_{F}(x)$ is measurable.
I never used the fact that $E \cup F = \mathbb{R^n}$, is what I've done still correct?
If it asks you to prove it by definition, it is probably expected that you show $f^{-1}(B)$ is Borel-measurable for any Borel-measurable set $B \subseteq \mathbb{R}^n$.
Note that since $E \cup F = \mathbb{R}^n$, we have the disjoint union
$$\mathbb{R}^n = (E \setminus F) \cup (E \cap F) \cup (F \setminus E)$$
Thus we have
$$f(x) = \begin{cases} 6, & \text{if $x \in E \setminus F$} \\ 1, & \text{if $x \in E \cap F$} \\ -5& \text{if $x \in F \setminus E$} \\ \end{cases}$$
So
$$f^{-1}(B) = \begin{cases} E \setminus F, & \text{if $B \cap \{6,1,-5\} = \{6\}$} \\ E, & \text{if $B \cap \{6,1,-5\} = \{6,1\}$} \\ (E \setminus F) \cup (F \setminus E)& \text{if $B \cap \{6,1,-5\} = \{6,-5\}$} \\ \mathbb{R}^n, & \text{if $B \cap \{6,1,-5\} = \{6,1,-5\}$} \\ E \cap F & \text{if $B \cap \{6,1,-5\} = \{1\}$} \\ F, & \text{if $B \cap \{6,1,-5\} = \{1,-5\}$} \\ F \setminus E& \text{if $B \cap \{6,1,-5\} = \{-5\}$} \\ \emptyset& \text{if $B \cap \{6,1,-5\} = \emptyset$} \\ \end{cases}$$
All the possible preimages are Borel-measurable so $f^{-1}(B)$ must be Borel-measurable.
We conclude that $f$ is measurable.