I want to show that the set of all Lebesgue measurable functions is a vector space. Consider a measurable space $(X,\epsilon)$ and a measure $\mu$ on this space.
Specifically, I wish to show that
$\mathcal{L}_0 (\mu) = \{ f : \mathbb{R} \rightarrow \mathbb{R} : \text{f is measurable and } \lim_{t \rightarrow \infty} \mu (\{ {x\in X: f(x) \ge t} \}) = 0 \} $
is a vector space. It is a sub space of the vector space of measurable functions. Of course $0\in \mathcal{L}_0 (\mu) $, but I am having trouble showing that it is closed under linear combinations.
I would like to use that for $\alpha \in \mathbb{R}$ and $f,g \in \mathcal{L}_0 (\mu)$, using the common notation $ \{ f \ge t \} := \{ {x\in X: f(x) \ge t} \} $, we have that
$\{ \alpha f + g \ge t \} \subseteq \{ |\alpha f + g |\ge t \} \subseteq \{ |\alpha | | f | + | g |\ge t \} \subseteq \{ |\alpha | | f | \ge t \} \cup \{| g| \ge t \}$
As the measure is an increasing function, it follows that
$\lim_{t \rightarrow \infty} \mu (\{ \alpha f + g \ge t \}) \leq \lim_{t \rightarrow \infty} \mu (\{| \alpha | | f | \ge t \}) + \lim_{t \rightarrow \infty} \mu (\{ | g | \ge t \}) ) = 0 $,
provided that $|f| \in \mathcal{L}_0 (\mu)$, whenever $f \in \mathcal{L}_0 (\mu)$
My question is: Is this a reasonable approach, and if so, how do I show $|f| \in \mathcal{L}_0 (\mu)$, whenever $f \in \mathcal{L}_0 (\mu)$? I figure that if I can establish that, I can easily show that then $ |\alpha| |f| \in \mathcal{L}_0 ( \mu) $ as well.
As stated this is false. We can have $f \in \mathcal L_0(\mu)$ with $-f \notin \mathcal L_0(\mu)$. Example: let $f(x)=-|x|$. If, in the definition of $\mathcal L_0(\mu)$, you replace the inequality $f(x) \geq t$ by $|f(x)| \geq t$ then $\mathcal L_0(\mu)$ becomes a vector space, but you seem to know already how to prove this.