Prove that $E[X|X]=X$

Question

In my statistics notes , they used the following result in an exercise $E[X|X]=X$ , where $X$ is a normal random variable. They simply stated that if I know $X$, the expected value is $X$. How can I prove that more rigorously?

Answer 1

The solutions linked here already use the uniqueness, but another way to show this is to look at the sets

$$ E_n^+:=\Big \{ \omega: X(\omega)-\mathbb{E}[X\vert X] > \frac{1}{n} \Big\} \quad \text{and} \quad E_n^-:=\Big \{ \omega: X(\omega)-\mathbb{E}[X\vert X] < \frac{1}{n} \Big\}, $$

which are both measurable w.r.t $\sigma(X)$. Note that $ \{ X\neq \mathbb{E}[X\vert X] \}=E^+\cup E^- $, while

$$ E^+=\cup_{n}E_n^+ \quad \text{and} \quad E^-=\cup_{n}E_n^-. $$

Using condition $(2)$ of your definition, you can show that both $E^+$ and $E^-$ occur with probability $0$.