prove...

Quote:

Originally Posted by flower3

prove that if f is continuous on [a,b] with $f(x) \geq 0 \ , \forall x \in [a,b]$ prove that if g is strictly increasing on [a,b]
with $\int_a^b f . dg =0$ then f(x)=0 $\forall x \in [a,b]$

The proof is the same as for usual Riemann integral. Procede by contradiction: assume that $f$ is not identically zero on $[a,b]$ . Using continuity, prove that there exists $\epsilon>0$ and $a\leq u<v\leq b$ such that $f(x)\geq \epsilon$ when $x\in[u,v]$ . Then, justify the following: $\int_a^b f\, dg\geq \int_u^v f\, dg\geq \epsilon \int_u^v dg=\epsilon(g(v)-g(u))>0$ . And conclude.