While studying concepts of measurable functions, there was a theorem suggesting that if any measurable...

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 $\{x:f(x)\ne g(x)\}$ 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
$\{x:d(x)>a\}=\{\begin{array}{ll}\text{a null set}& a\ge 0\\ \text{a full set}& a<0\end{array}$
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 $\{x:f(x)\ne g(x)\}$ into $\{x:f(x)=g(x)\}$ , 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
$\{x:d(x)>a\}=\{\begin{array}{ll}\text{a full set}& a\ge 0\\ \text{a full set}& a<0\end{array}$
However, the later statement is actually not plausible at all. Is there any contradictory logic among here?
Appreicate as always.