Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal

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Theorem

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.

Suppose that $\mu \restriction_\GG$, the restriction of $\mu$ to $\GG$, is $\sigma$-finite.


Let $f, g: X \to \overline \R$ be $\GG$-measurable functions.

Suppose that, for all $G \in \GG$:

$\ds \int_G f \rd \mu = \int_G g \rd \mu$


Then $f = g$ $\mu$-almost everywhere.


Proof

First, assume that $f$ and $g$ are $\mu$-integrable.

Observe:

\(\text {(1)}: \quad\) \(\ds \set {f \ne g}\) \(=\) \(\ds \bigcup_{n \mathop = 1}^\infty \set {\size {f - g} \ge \dfrac 1 n }\)

For each $n \in \N_{>0}$:

\(\ds \map \mu {\set {f - g \ge \dfrac 1 n} }\) \(=\) \(\ds \int_{\set {f - g \ge \frac{1}{n} } } 1 \rd \mu\)
\(\ds \) \(=\) \(\ds n \int_{\set {f - g \ge \frac{1}{n} } } \dfrac 1 n \rd \mu\)
\(\ds \) \(\le\) \(\ds n \int_{\set {f - g \ge \frac{1}{n} } } \paren {f - g} \rd \mu\)
\(\ds \) \(=\) \(\ds n \paren {\int_{\set {f - g \ge \frac{1}{n} } } f \rd \mu - \int_{\set {f - g \ge \frac{1}{n} } } g \rd \mu}\)
\(\ds \) \(=\) \(\ds 0\) by hypothesis as $\set {f - g \ge \dfrac 1 n} \in \GG$

Swapping $f$ and $g$, we also have:

\(\ds \map \mu {\set {g - f \ge \dfrac 1 n} }\) \(=\) \(\ds 0\)

Thus:

\(\text {(2)}: \quad\) \(\ds \map \mu {\set {\size {f - g} \ge \dfrac 1 n} }\) \(=\) \(\ds \map \mu {\set {f - g \ge \dfrac 1 n} } + \map \mu {\set {g - f \ge \dfrac 1 n} }\)
\(\ds \) \(=\) \(\ds 0\)


Therefore:

\(\ds \map \mu {\set {f \ne g} }\) \(=\) \(\ds \map \mu {\bigcup_{n \mathop = 1}^\infty \set {\size {f - g} \ge \dfrac 1 n } }\) by $\paren 1$
\(\ds \) \(\le\) \(\ds \sum _{n \mathop = 1}^\infty \map \mu { \set {\size {f - g} \ge \dfrac 1 n } }\)
\(\ds \) \(=\) \(\ds 0\) by $\paren 2$

Hence the result, by definition of almost-everywhere equality.

$\Box$


Now, consider general $f$ and $g$.

Recall that $\mu \restriction_\GG$　is $\sigma$-finite.

That is, there is an exhausting sequence $\sequence {E_n}_{n\in\N} \subseteq \GG$ such that:

$\map \mu {E_n} < \infty$

We define $\sequence {A_n}_{n\in\N} \subseteq \GG$ by:

$A_n := E_n \cap \set { \size f \le n} \cap \set {\size g \le n}$

Then $\sequence {A_n}_{n\in\N}$ is also an exhausting sequence.

Let $f_n := f \chi_{A_n}$ and $g_n := g \chi_{A_n}$.

Then $f_n$ and $g_n$ are $\mu$-integrable so that:

$\forall n \in \N : \map \mu {\set {f_n \ne g_n} } = 0$

On the other hand:

$\set {f_n \ne g_n} = \set {f \ne g} \cap A_n$

Therefore:

\(\ds \map \mu {\set {f \ne g} }\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \mu {\set {f \ne g} \cap A_n }\) Measure of Limit of Increasing Sequence of Measurable Sets
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \map \mu {\set {f_n \ne g_n} }\)
\(\ds \) \(=\) \(\ds 0\)


$\blacksquare$


Also see


Sources

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