Consider the set $[0,1]$ and the function $f:[0,1]\rightarrow \mathbb{R}$ given by simply by $f(x)=x$. I believe this function $f$ is measurable isn't it? Where I am getting confused is when I consider the set $(-0.1,0.1)\subset \mathbb{R}$. I belive this is in the Borel algebra $B\left ( \mathbb{R} \right )$ of $\mathbb{R}$.
The problem is that $f^{-1}\left ((-0.1,0.1 \right ))\nsubseteq [0,1]$, so I don't see how the set $f^{-1}\left ((-0.1,0.1 \right ))$ could be in the the Borel algebra $B([-0.1,0.1])$. But that would mean that we have a set $f^{-1}\left ((-0.1,0.1 \right ))\in B\left ( \mathbb{R} \right )$ that is unmeasurable in $[0,1]$. But that would mean that f is NOT measurable on the set $[0,1]$. Where am I going wrong?
When you have a function $f\colon A\longrightarrow B$ and a set $S\subset B$,$$f^{-1}(S)=\{a\in A\,|\,f(a)\in S\}.$$In particular, $f^{-1}(S)\subset A$. So, $f^{-1}\bigl((-0.1,0.1)\bigr)\subset[0,1]$. And, yes, $f$ is measurable.