# Prove that $\int_a^x f\,dx=0$ for all $x\in [a,b]$ implies $\int_a^b fg\,dx=0$ for any integrable $g$.

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 $$\left(\int_a^b fg \,dx\right)^2\leq \left(\int_a^b f^2 \,dx \right) \left( \int_a^b g^2 \,dx\right)=0$$

• Are you aware that a Riemann integrable function is continuous almost everywhere? – copper.hat May 11 '20 at 6:15
• 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. – Howardli621 May 11 '20 at 6:23
• Perhaps you could approximate $g$ be a sequence of step functions? – copper.hat May 11 '20 at 6:27
• 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 $f$. – Paramanand Singh May 11 '20 at 10:06

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 $$f$$.

• To have this inequality, you need to know that $f \geq 0$. And we don't know this. – Paul May 11 '20 at 6:36
• Decompose $f = f^+ - f^-$ – Moon Bears-C- May 11 '20 at 6:37
• 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 – Moon Bears-C- May 11 '20 at 7:01
• To make your argument with the decomposition work you would have to show that $\int_a^b f^+(x) dx = 0$ and $\int_a^b f^-(x) dx = 0$. – Martin R May 11 '20 at 7:26
• @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? – Martin R May 11 '20 at 7:59

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 $$a = x_0 < x_1 < \ldots and arbitrary “tags” $$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 \, .$$

• 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 – Paramanand Singh May 11 '20 at 10:16

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.