When is $\mathcal{L}_\infty$ a vector space? Suppose that $X$ is a subset of a Hausdorff topological space, and $Z\subseteq X$ a member of the family of Borel subsets on $X$. Let $B(Z)$ the set of bounded functions on $Z$, equipped with the $\sup$ norm. Then we can easily prove that every Cauchy sequence in $B(Z)$ converges.
Now take $BM(Z)$ the set of real-valued, bounded, measurable functions defined on $Z$. The fact that the functions are measurable, gives a hint as to define a norm on this space that uses this measure. Indeed, suppose that $\mu$ such a (finite, signed) measure on $Z$; the essential supremum norm $\|f\|^*_\infty = \mathrm{ess}\sup |f|$ yields the equivalence class of measurable functions that are bounded $\mu-$almost everywhere. Call this space $\mathcal{L}_\infty(Z,B(Z),\mu)$.
Question:
Suppose that we take a function $f\in \mathcal{L}_\infty$. Then this function may or may not in general exist in $B(Z)$, for it may assume an infinite value at some point in $Z$. Then how can one be assured that $\mathcal{L}_\infty$ is a vector space?
 A: 
The essential supremum norm ... yields the equivalence class of measurable functions that are bounded μ−almost everywhere.

This sentence indicates confusion.  Each element of $\mathcal{L}_\infty$ is an equivalence class of measurable functions.  Once you are used to this, you can go back to abusing notation and terminology like everyone else, but for now let's be careful.  If $f$ is an essentially bounded function, let $[f]\in\mathcal{L}_\infty$ denote the equivalence class of $f$.  If $[f]$ and $[g]$ are in $\mathcal{L}_\infty$, define $[f]+[g]=[f+g]$.  If $[f]$ is in $\mathcal{L}_\infty$ and $a$ is in $\mathbb R$, define $a\cdot [f]=[af]$.
With these operations, $\mathcal{L}_\infty$ is a vector space.  Checking that the operations satisfy the axioms is straightforward, but first you need to know that the definitions of the operations make sense.  For example, to show that $+$ is "well-defined", you need to show:


*

*If $f$ is essentially bounded and $g$ is essentially bounded, then $f+g$ is essentially bounded.

*If $f_1=f_2$ a.e. and $g_1=g_2$ a.e., then $f_1+g_1=f_2+g_2$ a.e.


Both facts rely on the fact that a union of two sets of measure zero is a set of measure zero.  Scalar multiplication is similar but easier.
A: Let $\mu$ be a positive measure defined on a $\sigma$-algebra $\mathcal{A}$ of subsets of the set $X$. It is easy to prove that the set $\mathcal{M}(X, \mathcal{A}, \mu)$ of all $\mu$- measurable functions $f \colon X \rightarrow \mathbb{K} \in \{ \mathbb{R}, \mathbb{C}\}$ is a linear subspace of the space of all maps from $X$ to $\mathbb{K}$.
Moreover, we can prove that the space $\mathcal{L}_{\infty}(X, \mathcal{A}, \mu)$ of all essentialy bounded functions $f \colon X \rightarrow \mathbb{K}$ is a linear subspace of $\mathcal{M}(X, \mathcal{A}, \mu)$. Indeed, let $f_1, \ f_2 \in \mathcal{L}_{\infty}$ and let $M_1$ and $M_2$ be such positive numbers that $|f_1(x)| \leq M_1$ and $|f_2(x)| \leq M_2$ almost every where on $X$. To show that $f_1+f_2 \in \mathcal{L}_{\infty}$ it is enough to note that
$$\{ x \in X  \colon |f_1(x) + f_2(x)| > M_1+M_2\} \subset \{ x \in X \colon |f_1(x)| >M_1\} \cup \{ x \in X \colon |f_2(x)| >M_2 \},$$
then $\mu (\{ x \in X \colon |f_1(x) +f_2(x) | > M_1+M_2\})=0.$ Verification that $\lambda f \in \mathcal{L}_{\infty}$ is even easier.
