First MVT for integral calculus

Question

Ist MVT for integral calculus states that:If f(x) and g(x) are bounded and integrable in [a,b] and g(x) maintains same sign in it, then, integration f(x)g(x) from a to b = c integration g(x) from a to b, where, m<=c<=M, where, m is infimum f(x) and M is supremum f(x) in [a,b]. Is 1st MVT for integral calculus applicable if the function g(x) =0 at some point x in [a,b]? I am asking this because in the statement it says that g(x) must be either positive or negative for all x in [a,b], which means that g(x) must not be 0 for any x in [a,b]

Answer 1

Let us take for granted the MVT for a strictly positive function $g(x)$.

Then consider a function $h(x)$ with arbitrary signs (and possibly zeroes). Let $H$ be a lower bound, i.e.

$$\forall x\in[a,b]:h(x)-H>0.$$

Thus the MVT applies to $g(x):=h(x)-H$ and there exists $c$ such that $g(c)=h(c)-H=\overline{h(x)-H}=\overline{g(x)}$. Then by linearity of the average,

$$h(c)=\overline{h(x)}.$$

This proves that signs do not matter.