to show if f is measurable or not

Question

Given $X=\{1,2,3,4\}$ and the sigma-Algebra $S=\{\emptyset,\{1\},\{2\},\{1,2\},\{3,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\} $, I've got to check out whether the function $f:(X,S)\rightarrow (\mathbb{R}, \mathscr{B(\mathbb{R}))}$is measurable or not for

i) $f(x)=(x-3)^2$ and ii) $f(x)=|x-\frac{7}{2}|$

My solution would be, that both functions are not measurable, because you have that $\{0\}\in \mathscr{B(\mathbb{R})}$ but $f^{-1}(\{0\})=3$ for i) and $f^{-1}(\{0\})=\frac{7}{2}$. As $3 \notin S$ and $\frac{7}{2} \notin S$, they would be not measurable. However this is only true for i) but does not count for ii). ii) is measurable and I don't know where and why I am wrong with my argument??

Answer 1

You stated that $f^{-1}(\{0\})=\frac72$ in case ii, which is not true.

We have $f^{-1}(\{0\})=\varnothing\in S$.

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The image of $f$ in case ii is $I:=\{\frac52,\frac32,\frac12\}$ and this with

• $f^{-1}\left(\{\frac52\}\right)=\{1\}\in S$
• $f^{-1}\left(\{\frac32\}\right)=\{2\}\in S$
• $f^{-1}\left(\{\frac12\}\right)=\{3,4\}\in S$

Also $\varnothing\in S$ so from this we are allowed to conclude that $f^{-1}(A)=f^{-1}(A\cap I)\in S$ for every subset of $\mathbb R$, hence is measurable.