Let $f$ and $g$ are two non-negetive simple functions on $X$. Then show that the set $A$ belongs to $£$, where $A=\{x:f(x)>=g(x)\}$ and $£$ is the sigma algebra of subsets of $X$.
Also I stuck with a problem in my mind that can we say the sets $\{x∈X:f(x)>g(x)\}$ and $\{x∈X:f(x)=g(x)\}$ are also elements of $£$? I think no, as if $AUB∈£$ we can not say that both belongs to $£$. Please help me for both.
First show that if $f$ and $g$ are measurable then $h = f-g$ is also measurable.
Then the problem is reduced to showing $\{x: h(x) \geq 0\}$ is measurable, as it can be written as $$ \{x: h(x) \geq 0\} = \{x: h(x)<0\}^c$$ and the latter is $$\{x: h(x)<0\} = \bigcup_{n\geq 1} \left\lbrace x: h(x) < -\frac{1}{n}\right\rbrace$$ Note that each set $\left\lbrace x: h(x) < -\frac{1}{n}\right\rbrace$ is $h^{-1}\left( -\infty, -\frac{1}{n}\right)$ which is measurable if $h$ is measurable.
The same also hold true for other sets like $\{x: h(x) = 0\}$ or $\{x: g(x)>0\}$ by the same way of re-writing subsets.