Let $(X,\Omega,\mu)$ be a measure space($\Omega$ is the $\sigma-$algebra and $\mu$ is the measure).
Let $(Y,T)$ be a topological space.
A function $f:X \rightarrow Y$ is called measurable iff for every $O \in T$
$$ f^{-1} (O) \in \Omega $$
If $f$ and $g$ are two measurable functions, then is the set $E = \{ x \in X : f(x) = g(x) \}$ measurable ?
I have found counterexamples showing that it does not hold in case of general topological spaces. What I wish to know is what happens when $Y$ is a general metric space ?
Any help would be appreciated.
Saying $Y$ is a metric space and $T$ is the $\sigma$-algebra of Borel sets on $T$ ... that is not enough.
Counterexample.
Let $Y$ be a set with cardinal $|Y| > \mathfrak c$, where $\mathfrak c = 2^{\aleph_0}$. With the discrete metric, $d(y_1,y_2) = 1$ if $y_1 \ne y_2$, we get of course that $T = \mathcal P(Y)$, the power set of $Y$.
Consider the product $\sigma$-algebra $T \otimes T$ on the Cartesian product $Y \times Y$. From $|Y| > \mathfrak c$ we can prove that $T \otimes T$ is not the power set of $Y \times Y$. Indeed, the "diagonal" $$ \Delta = \{(y,y) : y \in Y\} $$ is not an element of $T \otimes T$. (see HERE .)
For a measurable function $f,g : X \to Y$, define $h : X \to Y \times Y$ by $h(x) = \big(f(x), g(x)\big)$. Then $$ \{x : f(x) = g(x) \} = h^{-1}\big(\Delta\big) . $$ Now $\Delta \not\in T \otimes T$, so it could happen that $\{x : f(x) = g(x) \} \not\in \Omega$ . The simplest way to get an example is to take $X = Y \times Y$ and $\Omega = T \otimes T$ and $h$ the identity function on $Y \times Y$. That is, $f\big((x_1, x_2)\big) = x_1$ and $g\big((x_1, x_2)\big) = x_2$.
To get the opposite answer, you could limit $Y$ to be a separabale metric space.
Another nice case is when $(X, \Omega, \mu)$ is a complete measure space.