Theorem 3

Tag 08LK, permanent. Cite it as [Bourbaki, Tag 08LK].

Integration, chapter III, § 1. Measures on a locally compact space

The space $\mathscr{M}(X; \mathbf{R})$ of real measures on a locally compact space $X$ is fully lattice-ordered.

This follows from Ch. II, §2, No. 2, Th. 1.

In conformity with the notations of Ch. II, §1, we define, for every real measure $\mu$ on $X$ ,

\mu^+ = \sup(\mu, 0),\quad \mu^- = \sup(-\mu, 0),\quad |\mu| = \sup(\mu, -\mu);

then $\mu = \mu^+ - \mu^-,\ |\mu| = \mu^+ + \mu^-$ and $\inf(\mu^+, \mu^-) = 0$ . Moreover, for every function $f \in \mathscr{K}_+(\mathrm{X})$ ,

\int f\, d\mu^+ = \sup_{0 \leq g \leq f,\ g \in \mathscr{K}(\mathrm{X})} \int g\, d\mu

and

\int f\, d|\mu| = \sup_{|g| \leq f,\ g \in \mathscr{K}(\mathrm{X})} \int g\, d\mu,

whence, in particular,

\left| \int f\, d\mu \right| \leq \int |f|\, d|\mu|

for every function $f \in \mathscr{K}(\mathrm{X}; \mathbf{R})$ .

This inequality is also true if $f \in \mathscr{K}(\mathrm{X}; \mathbf{C})$ ; for, on multiplying $f$ by a complex number of absolute value 1 (which does not change either side of the inequality), one can suppose that $\int f\, d\mu \geq 0$ . Then

\left| \int f\, d\mu \right| = \int f\, d\mu = \int (\Re f)\, d\mu \leq \int |\Re f|\, d|\mu| \leq \int |f|\, d|\mu|.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Nothing.

Theorem 3

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