For example, say $f(x),g(x)$ are functions that are equal to each other in measurable space $E$, except on a subset $N$. Say a given measure on ``disagreeable'' space $N$ is equal to zero. Now $N$ is like Las Vegas and we are given license to forget about what happens in this space $N$ and assert that $f(x)=g(x)$ a.e where a.e stands for almost everwhere. To be more precise (just to prevent extra point from being leaked out of your exam paper!), we need to state $f(x)=g(x)$ a.e[$\mu$]. That is we need to specify which measure.
Mathematically, \begin{equation} \mu\{x:f(x)\neq g(x)\} = 0 \end{equation}where $x \in N$. The above criteria specifies the points of $N$. Here $f$ ~ $g$ and it is not too difficult to prove that this is an equivalence relation.
Reflexive property: Clearly $f$ ~ $f$. The disagreeable set $N$ in this case is a null set and measure of null set is $0$. $f=f$ on the whole set. Symmetric: Clearly $f$ ~ $g$ also means that set of disagreeable space remains same when we switch $f$ to the right. Reflexive: Say $f$ ~ $g$ and let $N_1$ be the disagreeable set. Say $g$ ~ $h$ and let $N_2$ be the disagreeable set, then $N_1 \cup N_2$ where $f,g,h$ are not equal to each other. Hence $f$ ~ $h$.
Note all the above statements were made based on a property that $f=g$. Generally, we don't have to be specific about a property. Above assertions and proofs hold in a more abstract sense, that is for any property $P$.
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