Proof o measurability of functions using pre-image definition The title is self explanatory. I want to show that, given two measurable functions$^1$ $f,g: X_1 \rightarrow \mathbb{R}$, we have that:

*

*$\alpha f + \beta g$, given $\alpha, \beta \in \mathbb{R}$, are measurable;

*$f^+ \equiv \sup_{x \in X_1} \{f(x), 0\}$ is measurable;

*$|f|$ is measurable.


My attempt of solution:

*

*I defined $F \doteq \alpha f + \beta g$. Then, I've tried to show that $F^{-1}[A] \in \mathcal{A}_1$, $\forall A \in \mathcal{D}$, as follows:

$$F^{-1}[A] = (\alpha f + \beta g)^{-1} [A] = \alpha f^{-1}[A] + \beta g^{-1} [A]$$
Noting that both $f$ and $g$ are measurable functions, we have that $\alpha f^{-1}[A] \in \mathcal{A}_1$ and that $\beta g^{-1} [A] \in \mathcal{A}_1$, $\forall A \in \mathcal{F}$. My problem is in the next step and is the following: given what I've said about $f$ and $g$, can I conclude that $\alpha f^{-1}[A] + \beta g^{-1} [A]\in \mathcal{A}_1$ or this is not aways true? If it is true, then I guess the proof os done;


*I don't know how to start this one, to be honest;


*Neither this one.
So, any tip for any of the three itens would be very appreciated.

$^1$ The definition I'm using is the follow: let $(X,\mathcal{A})$ be a measurable space, where $X$ is a set and $\mathcal{A}$ a $\sigma$-algebra over $X$. A function $f:X_1 \rightarrow \mathbb{R}$ is said to be a measurable function if $f^{-1}[A] \in \mathcal{A}$, $\forall A \in \sigma(\mathbb{R})$.
 A: First you have to assume that the domain $X_{1}$ is measurable. (you don't say it).
Then you only have to use Royden's Proposition 5.19.
A set is measurable if {$x:f(x)>c$}, or {$x:f(x)<c$} or {$x:f(x)\leq\,c$}, or {$x:f(x)\geq\,c$} is measurable for all $c\in\,\mathbb{R}$.
If $a=0$ then $af=0$ and is clearly measurable.
If $a>0$ then {$x:af(x)>c$}={$x:f(x)>\dfrac{c}{a}$} which is measurable.
Likewise for $a<0$. Thus $af$ is measurable for all $a\in\,\mathbb{R}$.
Now we prove that $f+g$ is measurable. If $f(x)+g(x)<c$ then $f(x)<r<c-g(x)$ for a rational $r$. Then {$x:f(x)+g(x)<c$}=$\bigcup_{r}^{}${$x:f(x)<r$}$\bigcap${$x:g(x)<c-r$} and since rationals are countable this is a countable union of measurable sets and therefore measurable.
By these statements we clearly get $af+bg$ measurable for all $a,b \in\,\mathbb{R}$.
The statement $2$ is clearly incorrect because $\sup_{X_{1}}${$f(x),0$} is a number or $+\infty$ and not a function. You probably mean that $f^{+}$=max{$f(x),0$} is measurable. But $f^{+}(x)=f(x)$ on
$A$={$x:f(x)>0$} and $f^{+}(x)=0$ on {$x:f(x)\leq\,0$}.
Then $f^{+}(x)=f(x)\chi_{A}(x)$ which is clearly measurable. We can define the negative part $f^{-}(x)$=max{$-f(x),0$} which is obviously measurable and since $|f|=f^{+}+f^{-}$ then $|f|$ is measurable.
