Proposition 13.1.12. Let $(E, \le)$ be a vector lattice, then:
For any $x, y \in E$,
\[x + y = x \vee y + x \wedge y\]Let $x \in E$, $x^{+} = x \vee 0 \ge 0$, and $x^{-} = -(x \wedge 0) \ge 0$, then $x = x^{+} - x^{-}$ and $|x| = x^{+} + x^{-}$. Moreover, $(x^{+}, x^{-})$ are the unique disjoint non-negative elements of $E$ such that $x = x^{+} - x^{-}$.
For any $x, y \in E$ and $\lambda \in \real$,
$|\lambda x| = |\lambda| \cdot |x|$
$|x + y| \le |x| + |y|$.
Finally, for any $x, y \in E$ with $x, y \ge 0$,
$[0, x] + [0, y] = [0, x + y]$.
Proof. (1): By Proposition 13.1.2,
(2): By (1),
By Proposition 13.1.2 and Lemma 13.1.10,
and
Since $x^{+}, x^{-} \ge 0$, $|x^{+}| = x^{+}$ and $|x^{-}| = x^{-}$, so by (1),
so $x^{+} \perp x^{-}$.
Now, let $y, z \in E$ with $y, z \ge 0$ such that $x = y - z$, then
and $x^{-} \le z$. If $y \perp z$, then $y - x^{+} \perp z - x^{-}$. Since $y - x^{+} = z - x^{-}$, $y - x^{+} = z - x^{-} = 0$, so $y = x^{+}$ and $z = x^{-}$.
(3): For any $\lambda > 0$, by (LO2),
(4): For any $x, y \in E$, by Proposition 13.1.2,
Likewise, $x^{-} + y^{-} \ge (x + y)^{-}$. Thus
(5): Let $x, y \in E$ with $x, y \ge 0$, then $[0, x] + [0, y] \subset [0, x + y]$. For any $z \in [0, x + y]$, let $u = z \wedge x$ and $v = z - u$, then