Lebesgue measure/Measurable sets Question :
Let $f,g$ be measurable real valued functions on $\mathbb{R}$ such that  :
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx$$
Let $E=\{x\in \mathbb{R} : f(x)\neq g(x)\}$ . Which of the followng statements are necessarily true?

*

*$E$ is empty set

*$E$ is measurable

*$E$ has lebesgue measure $0$

*For almost all $x\in \mathbb{R}$ we have $f(x)=0$ and $g(x)=0$
Explanation: What all I could see is that second bullet and third bullet are probably correct. Because :
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx$$
i.e.,
$$\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx-2\int_{-\infty}^{\infty} f(x)g(x)dx=0$$
i.e.,
$$\int_{-\infty}^{\infty}(f(x)-g(x))^2dx=0$$
Though I have negative limits my function $(f(x)-g(x))^2$ is positive
So, I would see that $E=\{x\in \mathbb{R} : f(x)\neq g(x)\}$ is measurable and has measure $0$
Please tell me if what I have done is sufficient/clear.
 A: Yes, I think what you have done is right, provided that f and g are square integrable. And the last statement that f and g are zero is trivially false.
A: OP's explanation for second and third bullets are no-doubt correct but by giving a counter-example, we shall also have to explain why first and fourth bullets are false. Here I am going to provide it.
For the first bullet, let $~f(x)=5,~~g(x)=\begin{cases} 
      5 &\text{if}~~~~ x\ne 0 \\
      1 &\text{if}~~~~ x= 0
         \end{cases}$
Clearly $~\int_{-\infty}^{\infty} (f(x)^2+g(x)^2)dx=2\int_{-\infty}^{\infty} f(x)g(x)dx~$ and $~0\in E\implies E=\phi~.$hence first bullet is not correct.
For the fourth bullet, from the above example, as neither $~f(x)=0~$ nor $~g(x)=0~,$  we can conclude that the fourth bullet is also not correct.

Note: I know that I am attempting to answer an old question. But I think my answer will also useful for solving the problem to the future reader. So if you downvote my answer, please tell me, where I am wrong. Thank you.
