Show that the sum of measurable simple functions are measurable. Show directly that if $f,g$ $ : X\rightarrow \textbf{R}$ are $\mathfrak{A}$-measurable simple functions, then $f+g$ is $\mathfrak{A}$-measurable.   
My approach: Show that $\{x \in X: (f+g)(x) \leq \gamma \} \in \mathfrak{A}$, where $\gamma$ is $\sup_{x} f(x) + sup_{x} g(x)$. Since both $f$ and $g$ are measurable, the sets $\{ x \in X: f(x) \leq \sup_{x} f(x)\}$ and $\{x \in X: g(x) \leq sup_{x} g(x) \}$ belongs to $\mathfrak{A}$, hence both  $\{ x \in X: f(x) \leq \sup_{x} f(x) + \sup_{x} g(x)\}$ and $\{x \in X: g(x) \leq sup_{x} g(x) + \sup_{x} f(x) \}$ belongs to $\mathfrak{A}$ and then you can show $\{x \in X: (f+g)(x) \leq \gamma \} \in \mathfrak{A}$ which gives that $f + g$ are measurable.
Does my approach make sense?

What you do does not work. For example, $\{ x \in X: f(x) \leq \sup_{x} f(x)\}$ is simply $X$. Here is what works:
If $f(x)+g(x)<r$, then there are rational numbers $p$ and $q$ such that $f(x)<p$, $g(x)<q$ and $p+q<r$. If $$P=\{(p,q)\in\mathbb{Q}^2:p+q<r\},$$
then $$\{x\in X:f(x)+g(x)<r\}=\bigcup_{(p,q)\in P}f^{-1}\Big((-\infty,p)\Big)\cap g^{-1}\Big((-\infty,q)\Big).$$
Since $P$ is countable, this set is measurable.
