# If $\mathcal{S}=\{\emptyset, X\}$ the only $\mathcal{S}$-measurable functions from $X\to\mathbb{R}$ are the constant functions

I have proved the following statement and I would like to know if there are any gaps in my reasoning, thank you:

"Suppose $$(X,\mathcal{S})$$ is a measurable space (i.e. $$X$$ is a set and $$\mathcal{S}$$ is a $$\sigma$$-algebra on $$X$$).

We say that a function $$f:X\to\mathbb{R}$$ is $$\mathcal{S}$$-measurable if $$f^{-1}(B)\in\mathcal{S}$$ for every Borel set $$B\subset\mathbb{R}$$."

If $$\mathcal{S}=\{\emptyset, X\}$$ then the only $$\mathcal{S}$$-measurable functions from $$X\to\mathbb{R}$$ are the constant functions".

My proof:

Let $$X$$ be a set and $$f:X\to\mathbb{R}$$ be a non-constant measurable function: then there will be $$y_1,y_2\in\mathbb{R}, y_1\neq y_2$$ such that $$y_1, y_2\in f(X)$$ (and there will also be $$x_1,x_2\in X, x_1\neq x_2$$ such that $$f(x_1)=y_1$$ and $$f(x_2)=y_2$$). Now, $$f^{-1}( \{y_1\} )=\mathcal{B_1}$$ and $$f^{-1}(\{y_2\})=\mathcal{B_2}$$, where $$\mathcal{B_1}$$ and $$\mathcal{B_2}$$ are Borel sets. There are four possible cases: either $$\mathcal{B_1}=\mathcal{B_2}=\emptyset$$, only one of them is $$\emptyset$$ and the other one is $$X$$ or $$\mathcal{B_1}=\mathcal{B_2}=X$$.

Suppose that one of $$\mathcal{B_1}$$ or $$\mathcal{B_2}$$, say $$\mathcal{B_1}$$, were the empty set: then $$x_1\in\mathcal{B_1}=\emptyset$$, a contradiction: thus the first three cases are impossible.

Suppose now that $$\mathcal{B_1}=\mathcal{B_2}=\emptyset$$: then $$f(x_1)=f(x_2)=X$$ so $$y_1=y_2$$, a contradiction.

Thus $$f:X\to\mathbb{R}$$ cannot be non-constant.

If it were constant instead, $$f:X\to\mathbb{R}, f(x)=c$$ for every $$x\in X$$ then $$f^{-1}(B)=X\in\mathbb{S}$$ for every Borel set containing $$c$$ and $$f^{-1}(B)=\emptyset\in\mathcal{S}$$ for every Borel set which does not contain $$c$$ thus $$f^{-1}(B)\in\mathcal{S}$$ in every possible case.

We have thus showed that if $$\mathcal{S}={\emptyset, X}$$ then the only $$\mathcal{S}$$-measurable functions are the constant functions, as desired.

• I read over it quickly and it looks ok. May 18, 2021 at 18:43
• @QuantumSpace Thank you for your time. May 18, 2021 at 18:44
• It looks okay, but overly complicated. Since $y_1,y_2\in f(X)$, then $f^{-1}(\{y_1\})$ and $f^{-1}(\{y_2\})$ are nonempty. And since $f^{-1}(\{y_1\})\cap f^{-1}(\{y_2\})=f^{-1}(\{y_1\}\cap\{y_2\}) = f^{-1}(\varnothing) = \varnothing$, then $f^{-1}(\{y_i\})\neq X$. Thus, $f$ is not measurable. May 18, 2021 at 18:45

Suppose that $$f: X \to \mathbb{R}$$ is $$\mathcal{S}$$-measurable and consider $$y\in \operatorname{Im}(f)$$. Then $$\{x \in X: f(x) = y\} \in \mathcal{S}$$
but since $$y$$ is in the image of $$f$$, this set is non-empty. Hence, this set must be $$X$$ and thus $$f(x) = y$$ for all $$x \in X$$, i.e. $$f$$ is contant.