a) Let H be an increasing directed set of functions ⩾0 that are lower semi-continuous on every compact subset of T. Then
μ∙(h∈Hsuph)=h∈Hsupμ∙(h).
b) Let H be a decreasing directed set of functions ⩾0 that are upper semi-continuous on every compact subset of T. If there exists in H a function h0 such that μ∙(h0)<+∞, then
μ∙(h∈Hinfh)=h∈Hinfμ∙(h).
For every compact set K⊂T, we have in case a)
μ∙(h∈HsuphφK)=μK∙(h∈HsuphK)=h∈HsupμK∙(hK)=h∈Hsupμ∙(hφK),
and in case b)
μ∙(h∈HinfhφK)=μK∙(h∈HinfhK)=h∈HinfμK∙(hK)=h∈Hinfμ∙(hφK),
by Prop. 2 of No. 2, and Prop. 8 of Ch. V, §1, No. 2. Case a) follows at once, by passage to the supremum with respect to K (No. 2, Prop. 2). To treat case b), denote by ε a number >0, and choose a compact set K such that μ∙(h0φK)⩾μ∙(h0)−ε. We then have (No. 5, Prop. 4) μ∙(h0φCK)⩽ε; for every function h∈H that is ⩽h0, we therefore have μ∙(hφCK)⩽ε, and finally μ∙(hφK)⩾μ∙(h)−ε by Prop. 4 of No. 5. Therefore
μ∙(h∈Hinfh)⩾μ∙(h∈HinfhφK)=h∈H,h⩽h0infμ∙(hφK)⩾h∈H,h⩽h0infμ∙(h)−ε.
Consequently the left side of (6) is ⩾ the right side; the reverse inequality being obvious, the proposition is established.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.