Let $g:[0,1]\to\mathbb{R}$ be bounded and measurable. For every continuous function $f$ with $\int_0^1f(x)dx=0$, $\int_0^1f(x)g(x)dx=0$ holds.
I want to prove that $g$ is a constant function on almost everywhere.
I approached in various way:
- I defined a function $h(c)=m(\{x\ \vert\ \vert g(x)-c\vert>0\})$ and tried to prove the existence of a root, but it was difficult to approach.
- I consider new function $g_0=g+\vert\vert g\vert\vert_\infty$ and I found that I can assume $g$ be positive because $g_0$ satisfies all the properties of $g$ what I need.
- Since $[0,1]$ is closed and $f$ is continuous, we can know that $f$ is bounded on $[0,1]$. So, we can define $f_0$ in the analogous sense with what I done to $g$.
But I can not do further. Please give me a hint or the concepts I should know. Any advice is welcome. Thanks.
Hint: What is $\int_0^1 f(x) g(x)\; dx$ when $f$ is an arbitrary continuous function?