Show that there exists an $\sum$-measurable simple function $\phi$ such that: $\int |f-\phi| d\mu <\epsilon$

Question

Problem:

Let $f \in L(X;\Sigma)$ where $L(X;\sum)$ is the set of integrable functions that can be written as $f=f^{+}-f^{-}$ where $\int f^{+} d\mu < \infty $ and $\int f^{-} d\mu < \infty $ where $f^{+}=max{f(x),0}$ and $f^{-}=max{-f(x),0}$.

Let $\epsilon>0$. Show that there exists an $\sum$-measurable simple function $\phi$ such that: $\int |f-\phi| d\mu <\epsilon$

The first two lines in the book's solution are:

We write $f=f^{+}-f^{-}$. Let $A = \{x\in X : f^{+}(x) = 0\}$ and $B = \{x\in X : f^{-}(x) = 0\}$. Then $X=A\cup B $.

Can anyone, please, explain to me why is $X=A\cup B $ ? I don't get this part at all. For example: the set $C = \{x\in X : f^{+}(x) > 0\}$ also belongs to $X$ but NOT to $A\cup B$. I don't think $A\cup B$ covers the whole $X$. Thanks!

Answer 1

For every $x \in X$, we have three possibilities:

1. $f(x) > 0$, so $f^+(x)$ is nonzero and $f^-(x)$ is zero, hence $x \in B$ and $x \not\in A$
2. $f(x) < 0$, so $f^-(x)$ is nonzero and $f^+(x)$ is zero, hence $x \in A$ and $x \not\in B$
3. $f(x) = 0$, so both $f^-(x)$ and $f^+(x)$ are zero, hence $x \in A$ and $x \in B$

In all three cases, $x$ is in either $A$ or $B$ or both. Thus $X \subseteq A \cup B$. The reverse containment is clear, so $X = A \cup B$.

Note that your set $C = \{x \in X : f^+(x) > 0\}$ is in fact the same as $B$, since $f^+(x) > 0$ if and only if $f(x) > 0$. Thus $C = B \subseteq A \cup B$.