Let $(X,\mathcal{A},\mu)$ be a measure space. A set $N\subseteq X$ is said a null set if $N\in\mathcal{A}$ and $\mu(N)=0.$ We denote with $\mathcal{N}_\mu$ the collection of the null sets.
A set $E\subseteq X$ is said negligible if exists $N\in\mathcal{N}_\mu$ such that $E\subseteq N$. We denote with $\mathcal{T}_\mu$ the collection of the negligible sets.
Let $(X,\mathcal{A},\mu)$ be a measure space. Two functions $f,g\colon X\to \overline{\mathbb{R}}$ they say equal almost everywhere in $X$ if $$\{x\in X\;|\; f(x)\ne g(x)\}\in\mathcal{N}_\mu.$$
It can be shown that a measure space $(X,\mathcal{A},\mu)$ is complete iff $\mathcal{N}_\mu=\mathcal{T}_\mu$.
Proposition. Let $(X,\mathcal{A},\mu)$ a complete measure space. Then the a.e. equality is an equivalence relation in $\overline{\mathbb{R}}^X$.
Proof. Transitivity. Let $f,g,h\colon X\to \overline{\mathbb{R}}$ such that $f=g$ a.e, and $g=h$ a.e. We place $$N_1=\{x\in X\;|\; f(x)\ne g(x)\}\in\mathcal{N}_\mu,\quad N_2=\{x\in X\;|\; g(x)\ne h(x)\}\in \mathcal{N}_\mu.$$ We define $N=N_1\cup N_2$, then $N\in\mathcal{N}_\mu$. Since $N^c=N_1^c\cap N_2^c$, for each $x\in N^c$ we have $f(x)=g(x)=h(x)$. Then $$N^c\subseteq \{x\in X\;|\; f(x)=h(x)\}\implies\{x\in X\;|\; f(x)\ne h(x)\}\subseteq N.$$ Since $\mu$ is complete $$\{x\in X\;|\; f(x)\ne h(x)\}\in\mathcal{N}_\mu,$$ then $f=h$ a.e.
Quesstion. This proposition hold also the measure space is not complete. How can I show this?
With your definition of a.e. equality, completeness is necessary to ensure this is an equivalence relation.
To illustrate, consider the set $X := \{-1,0,1\}$ and the $\sigma$-algebra $\Sigma := \{\varnothing,\{0\},\{\pm1\},X\}$. Define the measure $\mu:\Sigma\to\bar{\mathbb R}$ to just be the zero function, so that $\mathcal N_\mu=\Sigma$.
Now, define three functions $f,g,h:X\to\bar{\mathbb R}$ as follows: $f(x) := |x|$, $g(x) := 0$, and $h(x) := x$.
Now, $\{x\in X \mid f(x)\neq g(x)\} = \{\pm1\}$ and $\{x\in X \mid g(x)\neq h(x)\}=\{\pm1\}$, showing that $f=g$ and $g=h$ a.e.. However, $\{x\in X \mid f(x)\neq h(x)\} = \{-1\}$ is not $\mu$-measurable. Therefore, $f$ and $h$ are not equal $\mu$-a.e. by your definition of a.e. equality.