Question Mean Value Theorem for Integrals

Question

I made a proof of the fact $∫_{a}^{b}f(x)g(x)dx=f(c)∫_{a}^{b}g(x)dx$, where f is continuous and g integrable and non-negative and $c$ is a value $a<=c<=b$. In the proof is assumed that $∫_{a}^{b}g(x)dx$ is non-zero (positive).My questions is what happend if this integral is zero?

Answer 1

Some elements below in the case $\int g =0$, to prove that the equality still holds.

If $g$ is non-negative with a vanishing integral, then $g$ is equal to zero almost everywhere. Therefore $f.g$ is also equal to zero almost everywhere as $f$ is supposed to be continuous and $\int fg=0$.

Note: I’m using Lebesgue integral here.

And if you want to use Riemann integral, you can consider an upper sum $U$ for $\int g$ that can be as small as you desire as $\int g=0$. Then notice that $f$ is bounded, let say by $M$ on $[a,b]$ as being continuous on this interval. Based on that you’ll find an upper sum for $\int fg$ smaller than $MU$.

Answer 2

If $g(x)$ is non-negative and $\int_a^b g(x) dx$ is zero, then $g(x)$ has to be zero for all $x \in [a,b]$. Therefore $\int_a^b f(x) g(x) dx$ has to be zero. So you can pick any $c \in [a,b]$ and$$ f(c) \int_a^b g(x)dx = f(c) \cdot 0 = 0 = \int_a^b f(x) g(x) dx. $$