Bourbaki

Proposition 2

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

Integration, V, p. 3, § 1. Essential upper integral

Let f,g,hf, g, h be three elements of F+\mathscr{F}_+; if gg and hh are measurable, then

∫∙f(g+h) dμ=∫∙fg dμ+∫∙fh dμ. \int^\bullet f(g + h)\, d\mu = \int^\bullet fg\, d\mu + \int^\bullet fh\, d\mu.

One is immediately reduced to the proof of the analogous formula for the upper integral. Since f(g+h)=fg+fhf(g + h) = fg + fh (with the convention that 0⋅(+∞)=00 \cdot (+\infty) = 0), we have

∫∗f(g+h) dμ≤∫∗fg dμ+∫∗fh dμ; \int^* f(g + h)\, d\mu \leq \int^* fg\, d\mu + \int^* fh\, d\mu;

it remains to establish the reverse inequality. Let uu be a lower semi-continuous function such that u≥f(g+h)u \geq f(g + h). Set v=ug+hv = \frac{u}{g + h} on the set where g+h>0g + h > 0, and v=+∞v = +\infty on the set where g+h=0g + h = 0; then v≥fv \geq f and u≥v(g+h)u \geq v(g + h), whence

∫∗v(g+h) dμ≤∫∗u dμ \int^* v(g + h)\, d\mu \leq \int^* u\, d\mu

and consequently, vv being measurable (Ch. IV, §5, No. 6, Cor. 4 of Th. 5),

∫∗fg dμ+∫∗fh dμ≤∫∗vg dμ+∫∗vh dμ=∫∗v(g+h) dμ≤∫∗u dμ, \int^* fg\, d\mu + \int^* fh\, d\mu \leq \int^* vg\, d\mu + \int^* vh\, d\mu \\ = \int^* v(g + h)\, d\mu \leq \int^* u\, d\mu,

which implies the desired inequality since uu is arbitrary.

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

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Transcribed from the printed volume. Not translated.