In my textbook there is this claim:
In $(\mathbb{R},\mathcal{M},m)$ where m is the Lebesgue measure, continuous function are measurable. Indeed, for every $\alpha \in \mathbb{R}$ the set $\{x\in\mathbb{R}:f(x)>\alpha \}$ is an open set of $\mathbb{R}$.
How can I prove that that set is open (using the $\delta-\gamma$ definition) ?
Thanks !
As it is mentioned in the comments the statement is not necesserily true if $M$ is the sigma algebra of the Lebesgue measurable sets.
Now in general if you want to prove that the set $A=\{x|f(x)>a\}=f^{-1}(a,+ \infty)$ is open, then:
Let $x \in A$. We have that $f(x) \in (a,+ \infty)$ and $(a,+ \infty)$ is an open subset of $\mathbb{R}$ thus exists $\epsilon>0$ such that $(f(x)- \epsilon,f(x)+\epsilon) \subseteq (a, +\infty)$
Also because of the continuity of $f$ exists $\delta>0$ such that $$f((x- \delta,x+ \delta)) \subseteq (f(x)-\epsilon,f(x)+\epsilon) \subseteq (a,+\infty)$$
Therefore $$(x-\delta,x+ \delta) \subseteq f^{-1}(a,+ \infty)=A$$
Proving that $A$ is open.