Integrable Function with Zero Integral on Sub-Sigma-Algebra is A.E. Zero/Proof 2
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.
Let $f : X \to \overline \R$ be a $\GG$-integrable function.
Suppose that, for all $G \in \GG$:
- $\ds \int_G f \rd \mu = 0$
Then $f = 0$ $\mu$-almost everywhere.
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:
\(\ds \int f^+ \rd \mu\) | \(=\) | \(\ds \int f \cdot \chi _{ G_+ } \rd \mu\) | $f^+$ is positive part of $f$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \int _{ G_+ } f \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | by hypothesis |
On the other hand, from $f \cdot \chi _{ G_- } \le 0$ follows:
\(\ds \paren {f \cdot \chi _{ G_- } } ^+\) | \(=\) | \(\ds 0\) |
and:
\(\ds \paren {f \cdot \chi _{ G_- } } ^-\) | \(=\) | \(\ds - \paren {f \cdot \chi _{ G_- } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren { -f } \cdot \chi _{ G_- }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f^-\) | $f^-$ is negative part of $f$ |
Thus:
\(\ds \int _{ G_- } f \rd \mu\) | \(=\) | \(\ds \int f \cdot \chi _{ G_- } \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int \paren {f \cdot \chi _{ G_- } } ^+ \rd \mu - \int \paren {f \cdot \chi _{ G_- } } ^- \rd \mu\) | integral of $f \cdot \chi _{ G_- }$ | |||||||||||
\(\ds \) | \(=\) | \(\ds - \int f ^- \rd \mu\) |
which implies:
\(\ds \int f ^- \rd \mu\) | \(=\) | \(\ds - \int f \cdot \chi _{ G_- } \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - \int _{ G_- } f \rd \mu\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | by hypothesis |
Therefore:
- $\ds \int \size f \rd \mu = \int f^+ \rd \mu + \int f^- \rd \mu = 0$
$\blacksquare$