Two functions agreed to differ at points on a full set While studying concepts of measurable functions, there was a theorem suggesting that if any measurable function is altered on a null set, its measurability still remains.

If $f:E\to\mathbb R$ is measurable, $E\in\mathcal M,  g:E\to\mathbb R$ is such that the set $\left\{x:f(x)\ne g(x)\right\}$ is null, then $g$ is measurable.

They showed the following proof.
Consider the difference $d(x)=g(x)-f(x)$. It is zero except on a null set, so
$$\left\{x: d(x)\gt a\right\}=
 \begin{cases} 
\text{a null set} & a\ge0 \\
\text{a full set} & a\lt 0 \\
\end{cases}$$
Here, a full set is the complement of a null set. Since both null and full sets are measurable, $d$ is a measurable function. $g=f+d$ is thus measurable.
Now, I was curious whether the statement remains true if I change the set $\left\{x: f(x)\ne g(x)\right\}$ into $\left\{x: f(x)=g(x)\right\}$
, i.e., differ at points in a full set, because there was no doubt if I alter the proof as the following.
The difference $d(x)$ is still measurable since
$$\left\{x: d(x)\gt a\right\}=
 \begin{cases} 
\text{a full set} & a\ge0 \\
\text{a full set} & a\lt 0 \\
\end{cases}$$
However, the later statement is actually not plausible at all. Is there any contradictory logic among here?
Appreicate as always.

Let $f=0, A\subseteq E$ be a non-measurable subset and $N$ is a null set with $N\cap A=\varnothing.$
Then we let $g(x)=
\begin{cases}
1,&x\in A\\
0,&x\in N\\
-1,&x\in E\setminus(A\cup N)\\
\end{cases}$.
Thus,$\{x:d(x)>0\}=A$ which is non-measurable. So here $d(x)$ is not a measurable function.
