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Let $f$ be a (Riemann) integrable function such that $\displaystyle \int_a^x f\,dx=0$ for all $x\in [a,b]$.
Prove that $\displaystyle \int_a^b fg\,dx=0$ for any integrable $g$.

The question will be easy if we assume that $f$ is continuous, since the given condition will imply $f=0$ everywhere.
However, when $f$ is only required to be integrable, it can be some strange function, like the popcorn function. In this case, I have no idea how to start with, and I can only hope that the given condition will imply $\displaystyle \int_a^b f^2\,dx = 0$.
Then we can use the Cauchy inequality $$\left(\int_a^b fg \,dx\right)^2\leq \left(\int_a^b f^2 \,dx \right) \left( \int_a^b g^2 \,dx\right)=0$$

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  • $\begingroup$ Are you aware that a Riemann integrable function is continuous almost everywhere? $\endgroup$ – copper.hat May 11 '20 at 6:15
  • $\begingroup$ Yes, but I cannot figure out how it helps me to prove the proposition. Perhaps the set where $f \neq 0$ has measure 0 but I want a simpler proof. $\endgroup$ – Howardli621 May 11 '20 at 6:23
  • $\begingroup$ Perhaps you could approximate $g$ be a sequence of step functions? $\endgroup$ – copper.hat May 11 '20 at 6:27
  • $\begingroup$ I think it should be possible to prove that if $[c, d] $ is any subinterval of $[a, b] $ then $f$ vanishes at some point of $[a, b] $. This will imply the desired conclusion. However I am not able to show the property of $f$. $\endgroup$ – Paramanand Singh May 11 '20 at 10:06
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If $g$ is integrable in the Riemannian sense, then $g$ is bounded. Say $$ m \leq g(x) \leq M $$ for all $x \in [a,b]$. Then by monotonicity of the integral we have that

$$ m \int_a^b f(x) dx \leq \int_{a}^b f(x) g(x) dx \leq M \int_{a}^{b} f(x) dx. $$

Yet $$ \int_a^b f(x) dx = 0 $$ therefore $$ 0 \leq \int_a^b f(x) g(x) dx \leq 0. $$

As was pointed out in the comment below, this works for $f \geq 0$. To fix this write observe that $$ \int_a^b f(x) dx = 0 \Rightarrow \int_a^b f^2(x) dx = 0. $$ A proof of this fact is done below. Then run the above argument with $f^2(x)$ instead of just $f$.

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    $\begingroup$ To have this inequality, you need to know that $f \geq 0$. And we don't know this. $\endgroup$ – Paul May 11 '20 at 6:36
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    $\begingroup$ Decompose $f = f^+ - f^-$ $\endgroup$ – Moon Bears-C- May 11 '20 at 6:37
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    $\begingroup$ You can always decompose an integrable function as the difference of two non-negative functions. Take $f^+ = max(f(x), 0)$ and $f^- = max(-f(x),0)$. Then write $f = f^+ - f^-$. Since Integration is linear you can break it into two separate integrals, with each integrand being non-negative $\endgroup$ – Moon Bears-C- May 11 '20 at 7:01
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    $\begingroup$ To make your argument with the decomposition work you would have to show that $\int_a^b f^+(x) dx = 0$ and $\int_a^b f^-(x) dx = 0$. $\endgroup$ – Martin R May 11 '20 at 7:26
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    $\begingroup$ @MoonBears-C-: Will you address my above concern? Why does $ \int_a^x f\,dx=0$ for all $x$ imply that the integral for $f^+$ and $f^-$ is zero? $\endgroup$ – Martin R May 11 '20 at 7:59
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One can indeed show that $\int_a^x f(t)\,dt=0$ for all $x \in [a, b]$ implies that $\int_a^b f^2(x)\,dx = 0$ (which implies the desired conclusion, as you already noticed):

Assume on the contrary that $I = \int_a^b f^2(x)\,dx > 0$. It follows that for every sufficiently fine partition $a = x_0 < x_1 < \ldots <x_n = b$ and arbitrary “tags” $t_i \in [x_{i-1}, x_i]$ $$ \sum_{i = 1}^n f^2(t_i) (x_i - x_{i-1}) > \frac 12 I > 0 \, . $$ In particular there must be an interval $[x_{i-1}, x_i]$ such that $$ c = \inf \{ f^2(x) | x_{i-1} \le x \le x_i \} > 0 \, . $$ Then $$ \int_{x_{i-1}}^{x_i} f(t) \, dt \ge \sqrt c (x_i - x_{i-1}) > 0 $$ in contradiction to $$ \int_{x_{i-1}}^{x_i} f(t) \, dt = \int_a^{x_i} f(t)\,dt - \int_a^{x_{i-1}} f(t)\,dt = 0 \, . $$

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  • $\begingroup$ You found a nice way to handle things by using a very smart technique. The fact that $f^2$ is non-negative helps a lot. +1 $\endgroup$ – Paramanand Singh May 11 '20 at 10:16
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Let $F(x) =\int_{a} ^{x} f(t) \, dt$ so that $F(x) =0$ on whole of $[a, b] $. Next we observe that if $[c, d] $ is a sub-interval of $[a, b] $ then $f$ is continuous at some point $\xi\in[c, d] $ and thus $f(\xi) =F'(\xi) =0$ via Fundamental Theorem of Calculus.

Next let us assume that $\int_{a} ^{b} f(x) g(x)\, dx>0$ (the case of $<0$ can be handled by replacing $g$ with $-g$). Then there is sub-interval $[c, d] $ of $[a, b] $ of positive length on which $f(x) g(x) >0$. But this contradicts the fact that $f$ vanishes somewhere on this sub-interval. The contradiction proves the desired result.

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