Lebesgue measure/Measurable sets

Question

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.

Answer 1

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.

Answer 2

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.

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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.