Lemme 2...1
Soient \(f\in\mathcal{C}_{pm}^0([a,b],\mathbb{R}), g\in\mathcal{Esc}([a,b],\mathbb{R}) \text{ et } \varepsilon>0\).
On suppose que \(||f-g||_{\infty}\leq\varepsilon\).
On a alors : \(|\int_{[a,b]}f-\int_{[a,b]}g|\leq(b-a)\varepsilon\).
Demo 2...1:
On a : \(g-\varepsilon\leq f\leq g+\varepsilon\) et donc \(\int_{[a,b]}(g-\varepsilon)\leq I^- \text{ et }\int_{[a,b]}(g+\varepsilon)\geq I^+\)
On a alors : \(\int_{[a,b]}g-\varepsilon(b-a)=\int_{[a,b]}(g-\varepsilon)\leq I^-=\int_{[a,b]}f=I^+\leq\int_{[a,b]}(g+\varepsilon)=\int_{[a,b]}g+\varepsilon(b-a)\)
Donc on a bien : \(|\int_{[a,b]}f-\int_{[a,b]}g|\leq\varepsilon(b-a)\).