I'm trying to prove this:
If $f,g:(X,\mathcal{A}) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ are such that for any $a\in \mathbb{R}$, it holds that $\mu(\{x\in X: f(x)\leq a < g(x)\}) = 0$, then $\mu(\{x\in X: f(x) < g(x)\}) = 0$.
So far, I've come with the following sketch of proof but I'm not sure whether it is correct. Any suggestions or comments would be appreciated.
Proof: We first show that $\bigcup_{a\in \mathbb{R}}A_{a} = \{x\in X: f(x)< g(x)\}=:A$.
Then we take $a\in \mathbb{R}\setminus\mathbb{Q} $ and we define $A_a := \{x\in X: f(x)\leq a < g(x)\}$. By the density of $\mathbb{Q}$ in $\mathbb{R} $, we obtain that for all $x\in A_a$, there is $a'\in \mathbb{Q}$ such that $f(x)\leq a' < g(x)$.
Therefore there is a subset $S_a \subset \mathbb{Q}$ such that $A_a \subset\bigcup_{a'\in S_a}A_{a'}$. If we extend this to all $a \in\mathbb{R}\ $, then $A = \bigcup_{a\in \mathbb{R}}A_{a} \subset \bigcup_{a\in T}A_a$ where $T \subset \mathbb{Q}$.
Finally, by the countability of $\mathbb{Q}$, we can apply the sub-additive property of measures and we obtain that $\mu(A) \leq \sum_{a\in T}\mu(A_a)$