Almost sure equal probability

I have a problem that reads:

Let X be a random variable such that $X(\omega) = \omega$ and a random variable Y such that

$Y(\omega)$ = \begin{cases} \omega, & \text{if $\omega$ $\neq$ 1/2} \\ 2, & \text{if $\omega$ = 1/2} \\ \end{cases}

Then it provides the answer as

P($X(\omega)$ is not equal to $Y(\omega)$) = P(1/2) = 0.

P here refers to Lebesgue measure.

My question is how P($X(\omega) \neq$ $Y(\omega)$) = P(1/2) = 0 ? need some explanation on this.

• Both P($X(\omega) \neq$ $Y(\omega)$) and P(1/2) are incorrect. One should write $P[X\ne Y]$ or $P[\{\omega\mid X(\omega)\ne Y(\omega)\}]$, and $P[\{\frac12\}]$.
– Did
Commented Sep 23, 2013 at 5:49

The colloquial notion of the probability of the event $X(\omega)\neq Y(\omega)$ formally corresponds the Lebesgue measure (in this specific context) of the set $$\left\{\omega\in\mathbb{R}\,|\,X(\omega)\neq Y(\omega)\right\}.$$ What are the values of $\omega$ for which $\omega$ is in this set, that is, for which $X(\omega)\neq Y(\omega)$? Since $X(\omega)=\omega$ by definition, we're looking for such $\omega$ that $\omega\neq Y(\omega)$. But by the definition of $Y(\omega)$, there is only one such $\omega$: $\omega=1/2$. Hence, $$\left\{\omega\in\mathbb{R}\,|\,X(\omega)\neq Y(\omega)\right\}=\left\{\frac{1}{2}\right\}.$$ Then, the Lebesgue measure of any singleton is zero, yielding the desired result.
• "the event $X(\omega)\neq Y(\omega)$" does not exist. You probably mean the event $\{X\ne Y\}=\{\omega\in\Omega\mid X(\omega)\neq Y(\omega)\}$.
• I was being colloquial. I meant to say that when you translate the colloquial notion of the “event that $X(\omega)$ does not equal $Y(\omega)$” (in the grammatical sense of “event,” not the mathematical one) into a rigorous formal definition, you define this notion as the set $\{\omega\in\mathbb{R}\,|\,X(\omega)\neq Y(\omega)\}$. I'll rephrase it to better emphasize the distinction between being informal and being rigorous. Commented Sep 23, 2013 at 6:13