-1

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 !

José Carlos Santos
  • 427,504
Gmv93
  • 189
  • 3
    If $\mathcal{M}$ is the Lebesgue $\sigma$-algebra (i.e. the completion of the Borel $\sigma$-algebra) then this is not true, see https://math.stackexchange.com/questions/479441/example-of-a-continuous-function-that-is-not-measurable – Nigel Overmars May 15 '17 at 12:13
  • The set in question is $f^{-1}(\alpha,\infty)$. If $f:\mathbb{R}\to\mathbb{R}$ is continuous, and $U\subset \mathbb{R}$ is open, then you want to show that $f^{-1}(U)$ is open. This is a standard fact that should be in most basic real analysis/topology textbooks. – Prahlad Vaidyanathan May 15 '17 at 14:07

1 Answers1

1

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.

Marios Gretsas
  • 21,906