Show a set E is measurable iff its characteristic function is measurable.

I am trying to show that a set $E$ is measurable iff its characteristic function is measurable. I saw a proof online that someone wrote as a homework solution, however it doesn't make sense to me. They wrote:

Suppose $E$ is a measurable set in a measure space $X$. If $\alpha \leq 0$, then $\{x:\chi_E(x) < \alpha\}=\varnothing$, a measurable set. If $\alpha >1$ then $\{x:\chi_E(x)<\alpha\}=X$, a measurable set. Finally, if $0<\alpha \leq 1$, then $\{x:\chi_E(x)<\alpha\}=X-E$, a measurable set. We conclude that $\chi_E$ is a measurable function. Conversely, suppose that $\chi_E$ is a measurable function. Then $E=X-\{x:\chi_E(x)<1/2\}$ is a measurable set.

There is so much I do not understand about this proof.

1) I dont understand why when $0<\alpha \leq 1$ the set is equal to $X-E$, since the characteristic function will never take on values that are strictly greater than 0 and less than 1, will it?

2) I dont understand why in the first direction showing that $\{x:\chi_E(x)<\alpha\}$= a measurable set shows that $\chi_E$ is measurable.

3) I dont understand the converse direction.

Can someone either help me understand or suggest a more understandable proof? The way I thought I would need to prove it was by using what we know about measurable sets (using that a set, $E$, is measurable if $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)$ for all $A\in X$).

Solution 1:

1) If $0 < \alpha \leq 1$, then $\chi_{E}(x) < \alpha$ (note the strict inequality here) iff $\chi_E(x) = 0$.

2) This is just the definition of measurability for functions.

3) As above, $\chi_E(x) < 1/2$ iff $\chi_E(x)=0$. Therefore, $\{x: \chi_E(x) < 1/2 \} = \{x: \chi_E(x)=0 \} = X-E$ and $E = X - (X - E)$. The set $\{x: \chi_E(x) < 1/2 \}$ is measurable by definition, and therefore $X - E$ is measurable. Hence, $E$ is measurable because (countable) intersections and complements of measurable sets are measurable.

Solution 2:

Use the definition of measurability: a function $f : (X_1,S_2) \rightarrow (X_2,S_2)$, where $(X_1,S_2)$ and $(X_2,S_2)$ are $\sigma$-algebras, is measurable if for every $F \in S_2$ we have that $f^{-1}(F) = E \in S_1$. In other words, the pre-image of a measurable set will always be a measurable set.

Let $(X,S)$ be a $\sigma$-algebra. The characteristic function $\chi_E$ is a function $\chi_: (X,S) \rightarrow (\mathbb{R},S_L)$, where $(\mathbb{R},S_L)$ is the $\sigma$-algebra of Lebesgue-measurable sets over $\mathbb{R}$.

First, suppose $\chi_E$ is a measurable function. Let $F \in S$; we must now show that $\chi^{-1}(F) \in S$. We can write $F = F_1 \cup F_2 \cup F_3$, where

$$F_1 = \{ x \in F \mid x < 0\}$$ $$F_2 = \{ x \in F \mid 0 \leq x < 1\}$$ $$F_3 = \{ x \in F \mid x \geq 1\}$$

One or more of these three sets may be empty.

Clearly, $F_1$, $F_2$ and $F_3$ are all Lebesgue-measurable, since $F$ is. Moreover, $\chi^{-1}(F) = \chi^{-1}(F_1) \cup \chi^{-1}(F_2) \cup \chi^{-1}(F_3)$. Since all of the components of this union are measurable, and since the family of measurable sets $S$ is closed under union, $\chi^{-1}(F)$ is measurable. So let $F = [1,1]$; we have $\chi^{-1}(F) = E$, and therefore $E$ is measurable.

Conversely, suppose $E \in S$ is measurable. Suppose $F \in S_L$. We must show that $\chi^{-1}(F)$ is measurable. Again, we can write $F = F_1 \cup F_2 \cup F_3$. We have $$\chi^{-1}(F_1) = \emptyset ,$$ $$\chi^{-1}(F_2) = X \setminus E \text{ if } F_2 \neq \emptyset \text{ and } \emptyset \text{ otherwise},$$ $$\chi^{-1}(F_3) = E \text{ if } F_3 \neq \emptyset \text{ and } \emptyset \text{ otherwise}$$

Since $(X,S)$ is a $\sigma$-field, the family $S$ of measurable is closed under unions, so $F$ is measurable.