Is $X$ measurable from $F$ to Borel sets of $\mathbb{R}$?

Question

Let $X:\{0,1,2\} \rightarrow \mathbb{R}$ be $X(0)=X(1)=1$ and $X(2)=2$. $F=\{\emptyset, \{1\},\{0,2\},\{0,1,2\}\}$.

Is $X$ measurable from $F$ to Borel sets of $\mathbb{R}$?

My professor said this is an elementary question for his course. My question is what is the definition of a measurable function from one space to another space?

Answer 1

Given two measurable spaces $(X,\mathcal F)$ and $(Y,\mathcal E)$, we say that a function $f:X\to Y$ is measurable if, for all $H\in\mathcal E$, $f^{-1}[H]\in\mathcal F$.

A useful remark is that, in the special case where $Y=\Bbb R$ and $\mathcal E$ is the Borel $\sigma$-algebra, $f:X\to \Bbb R$ is measurable if and only if any (and all) of the following equivalent conditions holds:

1. for all $\alpha\in\Bbb R$, $\{x\in X\,:\, f(x)\le \alpha\}\in \mathcal F$
2. for all $\alpha\in\Bbb R$, $\{x\in X\,:\, f(x)\ge \alpha\}\in \mathcal F$
3. for all $\alpha\in\Bbb R$, $\{x\in X\,:\, f(x)< \alpha\}\in \mathcal F$
4. for all $\alpha\in\Bbb R$, $\{x\in X\,:\, f(x)> \alpha\}\in \mathcal F$

Some authors who are only concerned with real analysis might give this as definition of measurable function $(X,\mathcal F)\to\Bbb R$, without bouthering too much with introducing the subtleties of the category of measurable spaces.