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?

We can show this (not directly as you wish) using next theorem:
Theorem. Let $(X,\mathfrak{M})$ be measurable space, $Y$ is metric space and $F:\mathbb{R}^2 \to Y$ is continuous function. If $f,g:X \to \mathbb{R}$ are measurable function, then function $h:X \to Y: x \mapsto F(f(x),g(x))$ is measurable also.
In our example $Y=\mathbb{R}$ and $F(s,t)=s+t$.
