Suppose X is equipped with a measure and $\mathbb{R}$ is equipped with the Lebesgue measure. Define a preorder on the set of all measurable functions from $X$ to $\mathbb{R}$: $f\le g\Leftrightarrow \mu(\{x|f(x)>g(x)\})=0$.
Is this order complete (any subset that is bounded above has a least upper bound)?
In general, the order will not be complete. Consider $(X, \Sigma, \mu)$ where $X=[0,1]$, $\Sigma$ is the Lebesgue $\sigma$-algebra in $[0,1]$ and $\mu$ is the counting measure. Let $\Bbb R$ be equipped with the Lebesgue $\sigma$-algebra.
Let $\mathcal{F}$ be the set of all (Lesbesgue-Lebesgue) measurable functions from $X$ to $\Bbb R$. Note that, for all $f, g \in \mathcal{F}$ $$f\leq g\Leftrightarrow \mu(\{x|f(x)>g(x)\})=0 \Leftrightarrow \{x|f(x)>g(x)\}=\emptyset \Leftrightarrow \forall x \in X, f(x)\leq g(x)$$
Now, let $E$ be a non-measurable subset of $[0,1]$, that is a subset that is not in the Lebesgue $\sigma$-algebra. Let $$\mathcal{G}= \{ \chi_{\{x\}} : x \in E \}$$ where, for any set $A$, $\chi_A$ is the indicator function of $A$.
Clearly $\mathcal{G}$ is a subset of $\mathcal{F}$. Moreover, $\chi_{[0,1]} \in \mathcal{F}$ and, for all $f \in \mathcal{G}$, $f \leq \chi_{[0,1]}$. So $\mathcal{G}$ is bounded above in $\mathcal{F}$.
Suppose that there is $h \in \mathcal{F}$, such that $h$ is a least upper bound of $\mathcal{G}$. Then, since $h$ is an upper bound of $\mathcal{G}$ , we have $$E \subseteq \{x \in X : h(x) \geq 1 \}$$ Since $h$ is (Lesbesgue-Lebesgue) measurable, $\{x \in X : h(x) \geq 1 \}$ is in the Lebesgue $\sigma$-algebra. So, we have that $$E \subsetneq \{x \in X : h(x) \geq 1 \}$$ It means that there is $p \in \{x \in X : h(x) \geq 1 \}$ such that $p\notin E$. Let $\hat h$ be the function defined as: $\hat h(p)=0$ and $\hat h(x)=h(x)$ if $x \in X \setminus \{p\}$. It follows immediately that: $\hat h$ is an upper bound of $\mathcal{G}$, $\hat h \leq h$ and $h \not \leq \hat h$. Contradiction.
So, there is no least upper bound of $\mathcal{G}$ in $\mathcal{F}$.
Remark: In fact, the example above works for any $\sigma$-algebra $\Sigma$ that contains the singletons and is not $2^{[0, 1]}$. For instance the countable-cocountable $\sigma$-algebra.