Let f,g,h be three elements of F+; if g and h are measurable, then
∫∙f(g+h)dμ=∫∙fgdμ+∫∙fhdμ.
One is immediately reduced to the proof of the analogous formula for the upper integral. Since f(g+h)=fg+fh (with the convention that 0⋅(+∞)=0), we have
∫∗f(g+h)dμ≤∫∗fgdμ+∫∗fhdμ;
it remains to establish the reverse inequality. Let u be a lower semi-continuous function such that u≥f(g+h). Set v=g+hu on the set where g+h>0, and v=+∞ on the set where g+h=0; then v≥f and u≥v(g+h), whence
∫∗v(g+h)dμ≤∫∗udμ
and consequently, v being measurable (Ch. IV, §5, No. 6, Cor. 4 of Th. 5),
∫∗fgdμ+∫∗fhdμ≤∫∗vgdμ+∫∗vhdμ=∫∗v(g+h)dμ≤∫∗udμ,
which implies the desired inequality since u is arbitrary.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.