Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal
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
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $10.14 \ \text{(ii)}$