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Is this result true? If $g$ is a nonnegative function in $\mathcal{R}[a, b]$ with $\int_{a}^b g=0$, and $f$ is continous on $[a,b]$, then $\int_a^b fg = 0$.

If $g \in \mathcal{R}[a, b]$, and $\int_{a}^b g=0$, then for every possible partition $\mathcal{P} = \{[x_{k-1}, x_k]\}_{k=1}^n$of $[a, b]$ the lower sum is zero. As $g$ is nonnegative, $\inf \{g(x)|x_{k-1} \leq x \leq x_k\} = 0$ for all $k = 1, \dots, n$.

I'm unsure of where to go from here.

Anu
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  • Basically, given that $g$ is not continuous, you can't say that $g=0$ on the intervall, however it is the case almost everywhere... And thus $fg$ is also null almost everywhere... – Martigan Jan 09 '18 at 12:50
  • Are you allowed to use the mean value theorem? It says direct that the integral will be $0$. – mickep Jan 09 '18 at 12:50
  • @mickep Unfortunately, I asked this question because of a proof I saw of the mean value theorem for integrals. It discussed the case where $\int_a^b g > 0$, I wanted to know how you show that the mean value theorem holds when $\int_a^b g = 0$. – Anu Jan 09 '18 at 12:53
  • @Martigan I got this part- If $fg$ is null almost everywhere, the lower sum of every partition is $0$. Because we know $fg$ is integrable as the product of integrable functions, we can say that $\int_a^b fg = 0$. What I'm not clear on in this- why is $g$ null almost everywhere? I'm guessing we do have to use the fact that $g$ is continuous almost everywhere to show this- but I don't know how. – Anu Jan 09 '18 at 12:58
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    @Anu If you have the Cauchy-Schwarz inequality you could also use that one to conclude that the integral is $0$. (I don't give an answer, since maybe that is also "forbidden") – mickep Jan 09 '18 at 13:08
  • @mickep Yes, can use Cauchy Schwartz. That's brilliant, thank you! Just to make sure I got that right: $0 \leq |\int_a^b fg| \leq \int_a^b|fg| \leq (\int_a^b |f|^2)^\frac{1}{2} (\int_a^b |g|^2)^\frac{1}{2}= 0$, where $g^2$ has integral zero as on every subinterval $[x_{k-1}, x_k] \subset [a, b], \inf_{x \in [x_{k-1}, x_k]}g^2 = 0$.

    Out of interest, is there another way to do this?

     – Anu Jan 09 '18 at 13:54

1 Answers1

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First note that $fg $ is Riemann integrable as product of Riemann integrable functions. Since $|f|$ is continuous on $[a,b]$, it has maximum, hence \begin{align*} \left|\int_a^b fg\right| &\leq\int_a^b |fg|\\ &=\int_a^b |f|g&&\text{because $g$ non negative}\\ &\leq\max_{[a,b]}|f|\int_a^b g&&\text{because $f$ continuous on $[a,b]$}\\ &=0 \end{align*} thus $\int_a^b fg=0$.

Actually, the continuity of $f$ is not needed. The conclusion holds for any Riemann integrable $f$ because it can be show (with a bit of measure theory) that $g=0$ almost everywhere (proof).

Fabio Lucchini
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