$f,g$ are both measurable on a set $ \Omega $, can $ \{ x\in \Omega: f(x)=g(x) \} $ be non-measurable? Suppose $f,g$ are both measurable on a set $ \Omega $, can $ \{ x\in \Omega: f(x)=g(x) \} $ be non-measurable? 
My attempt: Let $\Omega$ be an open interval in $\Bbb{R}$, then it has a non-measurable subset, say $E$. Let $f=g$ on $E$, and $f>g$ one $\Omega \backslash E$. Is this ok? Thank you! 
 A: The answer depends on the measurable spaces involved.  
Example 1: Let $(\Omega, \Sigma)$ be a measurable space and $f,g$ functions from $\Omega$ to $\mathbb{R}$ and let us consider the Borel $\sigma$-algebra in $\mathbb{R}$ 
Then, if $f$ and $g$ are measurable, then $h=f-g$ is measurable and we have that $$\{x\in\Omega : f(x)=g(x)\} = \{x\in\Omega : h(x)=0\}$$ is measurable. 
Example 2: Consider $([0,1], \Sigma)$, where $\Sigma$ is the trivial $\sigma$-álgebra, that is $\Sigma=\{\emptyset, [0,1]\}$. Let $f$ and $g$ be functions from $([0,1], \Sigma)$ to $([0,1], \Sigma)$ defined by
$$f(x)=x$$
and
$$g(x)=1-x$$
It is imediate that $f$ and $g$ are measurable functions, but 
$$\{x\in\Omega : f(x)=g(x)\} = \{\tfrac{1}{2}\}$$
is NOT measurable. 
A: If you are talking about real-valued functions, as it seems, then your set is just:
$\{ x \in \Omega: f(x)-g(x)=0\} =h^{-1}(\{0\})$
where $h=f-g$ is a measurable function.
A: Let $X$ be a Hausdorff topological space with a countable base for open sets.
Let $f, g$ be measurable maps $\Omega \rightarrow X$ with respect to the $\sigma$-algebra of Borel sets of $X$, i.e. $f^{-1}(U)$ and $g^{-1}(U)$ are measurable for all open sets $U$ of $X$.
I claim that the set $A = \{x\in \Omega: f(x) = g(x)\}$ is measurable.
Define a map $h:\Omega \rightarrow X\times X$ by $h(x) = (f(x), g(x))$.
Let $U, V$ be open sets of $X$. Then $h^{-1}(U\times V) = f^{-1}(U)\cap g^{-1}(V)$.
This is a measurable set. Since $X$ has a countable base, every open set of $X\times X$ is a countable union of sets of the form $U\times V$, where $U$ and $V$ are open.
Hence $h$ is measurable with respect $\sigma$-algebra of Borel sets of $X\times X$.
Since $X$ is Hausdorff, the diagonal $\Delta = \{(x, x): x \in X\}$ is closed.
Hence $A = h^{-1}(\Delta)$ is measurable as claimed.
A: Edit. Taking advices from @ByronSchmuland, I modified the proof so that it takes a more direct approach.
Under a mild assumption on the target space, we can prove that $\{f = g\}$ is measurable.
Assume that $X$ is both a topological space and a measurable space that satisfies the following conditions:


*

*$X$ is a $T_0$-space, 

*$X$ is second countable with a countable base $\{ U_i \}_{i=1}^{\infty}$,

*The $\sigma$-algebra over $X$ contains all the Borel sets.


The first 2 conditions say that if $x, y \in X$, then $x = y$ if and only if for all $i$, either $x, y \in U_i$ or $x, y \notin U_i$. From this, for functions $f, g : \Omega \to X$, we can write
$$
\{ f = g \}
= \bigcap_{i=1}^{\infty} (\{ f \in U_i \} \cap \{ g \in U_i \}) \cup (\{ f \in X \setminus U_i \} \cap \{ g \in X \setminus U_i \}).
$$
When $f$ and $g$ are measurable, this set is a countable intersection of measurable sets hence is measurable. ////
Notice that these assumptions are automatically satisfied for $X = \Bbb{R}^n$.
For the case $X = \Bbb{R}$, the proof reads out with a slightly more friendly language: for real $x$ and $y$, $x = y$ if and only if for all rational $r \in \Bbb{Q}$, either $x, y > r$ or $x, y \leq r$. So we can write
$$ \{ f = g \} = \bigcap_{r \in \Bbb{Q}} (\{f > r\} \cap \{g > r\}) \cup (\{f \leq r\} \cap \{g \leq r\}), $$
which is measurable. ////
