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


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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:

$G_+ := \set {x \in X : \map f x > 0} \in \GG$

and:

$G_- := \set {x \in X : \map f x \le 0} \in \GG$

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:

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

$\blacksquare$

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