Show $\operatorname{Char}_{E}(x)$ is measurable iff $E$ is measurable

Question

As indicted in the title, i'm trying to prove that $\operatorname{Char}_{E}(x)$ is measurable iff $E$ is measurable, where $\operatorname{Char}_{E}(x)$ is the characteristic function on $E$, i.e. $\operatorname{Char}_{E}(x)$={$0$ if $x \notin E$ and $1$ if $x \in E$}.

So, let $\operatorname{Char}_{E}(x)$: $\mathbb{R} \rightarrow \mathbb{R}$. Let's assume that $\operatorname{Char}_{E}(x)$ is measurable, and therefore given any open set $U \subset \mathbb{R}$, we have $f^{-1}(U)$ is measurable. I need to use this to prove that E is measurable.

I'm quite new to measure theory, and so really i'm just brainstorming. But I want to do something like $\bigcap_{n=1}^{\infty}\left(1-1/n,1+1/n\right)$, an use the fact that the set of measurable functions $M$ is a $\sigma$-algebra to get a neighborhood of 1 such that it's preimage is $E$, however the neighborhood i've constructed through an infinite countable intersection will just be the set {$1$} and so will not be open. Need some serious help here!

Now I will assume that $E$ is measurable, and I want to show that given an arbitrary open set $U \subset \mathbb{R}$, $f^{-1}(U)$ is measurable in $\mathbb{R}$.

I'm just as lost trying to prove this converse, any insight is appreciated!!

Answer 1

Assume the function is measurable. We know that the set $$ \{x\in X |\mathbb{1}_E(x)>a \} $$ is measurable for any $a$, so that for $a=0$ we get that $E$ is measurable. Now if $E$ is measurable, then $$ \{x\in X |\mathbb{1}_E(x)>a \} $$ is clearly measurable for any $a$

Answer 2

All inverse images $f^{-1}(U)$ are among the sets: $$\mathbb R, \quad\varnothing,\quad E,\quad \mathbb R \setminus E.$$ We do not even have to require that $U$ is open for that. So it is enough to show that these four sets are measurable.

Answer 3

You have the right kind of idea but there isn't any need to take the intersection in the first part. Since $f = \operatorname{Char}_E$ only takes the values $0$ or $1$, we in fact have $E = f^{-1}(\frac12,\frac32)$ for example. Since $(\frac12,\frac32)$ is open, this shows that $E$ is measurable if $f$ is.

For the converse, if $U \subseteq \mathbb{R}$ is open (or even just measurable) then there are only four possiple preimages of $U$ and these are $\mathbb{R}, E, \mathbb{R} \setminus E$ and $\emptyset$ all of which are measurable if $E$ is.

Answer 4

Hint: Assume that $E$ is measurable, and let $U$ be a measurable subset of $\mathbb R$. There are only 4 relevant cases to consider: that is all combinations of whether $1$ or $0$ is in $U$. Thus there are four possibe preimages of $U$ under $\chi_E$, which are elucidated in other answers.