I've been given the following task
Let $(X,\mathscr{M}_X,\mu)$ a measure space. Two measurable mappings $f,g:X \to Y$ into a measurable space $(Y,\mathscr{M}_Y)$ are called equal almost everywhere if $$ N:= \{x \in X \mid f(x) \not= g(x)\}$$ is a null set.
Show that the almost everywhere equality is an equivalence relation "$\sim$".
So i've been working a bit on that proof and i am supposed to assume $Y$ is a Hausdorff space and $\mathscr{M}_Y = \mathscr{B}(\mathscr{O}_Y) = \mathscr{B}(\mathscr{A}_Y)$ in other words, the $\sigma-$Algebra $\mathscr{M}_Y$ on $Y$ is is the Borel-$\sigma-$Algebra generated by the open (or equivalently the closed) subsets of $Y$.
So my attempt would have been like this:
- $f \sim f$
The set $N_1 := \{x \in X \mid f(x) \not= f(x)\} = \emptyset$ is obviously measurable with $\emptyset \in \mathscr{M}_X$
- $f \sim g \Rightarrow g \sim f$
And here is my main issue. My first attempt was working on this step by simply assuming $$ N:= \{x \in X \mid f(x) \not= g(x)\}$$ is a null set.
However, as far as i'm concerned, i am supposed to prove that $N$ is measurable, i.e. $N \in \mathscr{M}_X$
That's where i'm a bit lost. I know that $Y$ is Hausdorff iff $$ \Delta Y := \{(y,y) \mid y \in Y\} $$ is closed in $Y\times Y$.
Therefore $(Y \times Y) \setminus \Delta Y$ is an open set. However, i have yet to figure out how to prove that $N \in \mathscr{M}_X$
Unfortunately i am stuck. Can anyone help me ?
Thank you very much for every help!
As $a \neq b$ iff $b \neq a$ for any numbers, $N= \{x: f(x) \neq g(x)\} = \{x: g(x) \neq f(x)\}$. It's just the same set, so if the former is a null set, so is the latter trivially. Nothing to prove: if $f \sim g$ by definition this is a null set.
For transitivity, assume $f \sim g$ and $g \sim h$. Note that $\{x: f(x) = g(x)\} \cap \{x: g(x)=h(x)\} \subseteq \{x: f(x)=h(x)\}$ so that (taking complements):
$$\{x: f(x)\neq h(x)\} \subseteq \{x: f(x)\neq g(x)\} \cup \{x: g(x) \neq h(x)\}$$
so that $\{x: f(x)\neq h(x)\}$ is a subset of the union of two null sets, hence a null set.
Hausdorffness or the precise $\sigma$-algebra's are irrelevant.