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Given a probability space $(\Omega ,\mathcal{F} ,\mu)$. Let $X$ and $Y$ be $\mathcal{F}$-measurable real valued random variables. How would one proove that $\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}$ is measurable.

My thoughts: Since $X$ and $Y$ are measurable, it is true, that for each $x\in\mathbb{R}:$ $\left\{ \omega|X(\omega)<x\right\} \in\mathcal{F}$ and $\left\{ \omega|Y(\omega)<x\right\} \in\mathcal{F}$.

It follows that $\left\{ \omega|X(\omega)-Y(\omega)\leq x\right\} \in\mathcal{F}$

Therefore $\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}$, for $n\in\mathbb{N}$.

Therefore $0=\bigcap_{n\in\mathbb{N}}\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}$.

Am working towards the correct direction? I appreciate any constructive answer!

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If you know that the difference of two measurable function is measurable, it's almost correct (in the last line, the LHS is the set $\{X=Y\}$, not $0$), but in this case you just have to show that $\{Z=0\}$ is measurable for $Z$ measurable. – Davide Giraudo May 2 '12 at 11:20
up vote 6 down vote accepted

$$[X\ne Y]=\bigcup_{q\in\mathbb Q}\left([X\lt q]\cap[Y\geqslant q]\right)\cup\left([X\geqslant q]\cap[Y\lt q]\right)=\bigcup_{q\in\mathbb Q}\left([X\lt q]\Delta[Y\lt q]\right) $$

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And, as mentioned by @Davide, the problem of the strategy in the post is that it assumes 90% of the result, namely the fact that if X and Y are measurable then Z=X-Y is measurable as well. – Did May 2 '12 at 12:48

Let $Z\colon \Omega\to\mathbb R^2$ defined by $Z(\omega)=(X(\omega),Y(\omega))$, where $\mathbb R^2$ is endowed with the Borel $\sigma$-algebra $\mathcal B(\mathbb R^2)$.

  • Show that the fact that the projections are measurable implies that so is $Z$.
  • Show that the set $\{(x,x)\in\Bbb R\times \Bbb R)$ is $\mathcal B(\mathbb R^2)$-measurable.
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