Measurable in functions (equal state) 
Let $f: \Bbb R\rightarrow \Bbb R $ with this properties:
$\forall \epsilon >0$ , $\exists U\subset\Bbb R$ , $U$ is open set,  s.t.$m(U)<\epsilon$ , $f$ is continuous on $ \Bbb R$\ $U$
How to prove that $f$ is measurable on $\Bbb R$.

in this question: measurable i.e. lebesgue-measurable.
we must be prove: $f$ is measurable i.e. if $\alpha $ is arbitrary then $\{ x \in \Bbb R | f(x) > \alpha \} $ is measurable.
$m :$ {measurable} $\rightarrow \Bbb R$, and $U$ is open set , so $m(U)$ is well-defined
 A: For every $n$, take an open set $U_n$ of measure $m(U_n)<1/n$ and such that $f$ is continuous on $\mathbb{R}\setminus U_n$. In particular, $f$ is measurable on $\mathbb{R}\setminus U_n$.
The set $V=\bigcap_{n=1}^\infty U_n$ has measure $m(V)=0$. It's complement is $\mathbb{R}\setminus (\bigcap U_n)=\bigcup(\mathbb{R}\setminus U_n)$. Since $f$ is measurable on each $\mathbb{R}\setminus U_n$, then $f$ is measurable on $\mathbb{R}\setminus V$ (for each borel set $W$, we have $f|_{\mathbb{R}\setminus V}^ {-1}(W)=\bigcup(f|_{\mathbb{R}\setminus U_n}^{-1}(W))$, a countable union of measurable sets, hence measurable).
Since $V$ has measure $0$ and $f$ is measurable on $\mathbb{R}\setminus V$, then $f$ is measurable (if $W$ is a borel set, then $f^{-1}(W)=(f|_{\mathbb{R}\setminus V}^{-1}(W))\cup(f|_V^{-1}(W))$. We already know that $f|_{\mathbb{R}\setminus V}^{-1}(W)$ is measurable, and $f|_V^{-1}(W)$ is a subset of $V$, hence has measure $0$ and thus it is measurable, since the Lebesgue measure is complete).
