Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero/Proof 2

Proof
In view of Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we shall show:
 * $\ds \int \size f \rd \mu = 0$

Since $f$ is $\GG$-measurable:
 * $G_+ := \set { x \in X : f(x) > 0 } \in \GG$

and:
 * $G_- := \set { x \in X : f(x) \le 0 } \in \GG$

On the one hand:

On the other hand, from $f \cdot \chi _{ G_- } \le 0$ follows:

and:

Thus:

which implies:

Therefore:
 * $\ds \int \size f \rd \mu = \int f^+ \rd \mu + \int f^- \rd \mu = 0$