Let $f$ be a real-valued Lebesgue measurable function on $\mathbb{R}$. Prove that there exist Borel functions $g$ and $h$ such that $g(x)=h(x)$ almost everywhere and $g(x)\le f(x)\le h(x)$ for every $x\in \mathbb{R}$.
My Solution:
I tried it as if $B\in \mathbb{B}(\mathbb{R})$ (where $\mathbb{B}(\mathbb{R})$ is Borel $\sigma - $algebra on $\mathbb{R}$) then $f^{-1}(B)=A\cup N_1$ where $A\in \mathbb{B}(\mathbb{R})$ and $\mu(N_1)=0$. Now i defined $g^{-1}(B)=A=h^{-1}(B)$.
now i am not able to understand what to do next like how to show $g(x)=h(x)$ almost everywhere and $g(x)\le f(x)\le h(x)$ for every $x\in \mathbb{R}$.
I would suggest another approach. Some hints: recall that every Lebesgue measurable positive function can be approximated by a (monotone) sequence of simple functions, and that every Lebesgue measurable set can be approximated by a Borel set.
To be a bit more precise (although not yet complete):
Without loss of generality, we assume that $f$ is non negative (otherwise, exploit the decomposition $f=f^{+}-f^{-}$); furthermore I for simplicity suppose that $f$ is bounded, say by $N \in \mathbb{N}$. Consider the pre-image sets: $$A^{i}_n=\{ x \in \mathbb{R} : \frac{i}{2^n} \le f(x) \le \frac{i+1}{2^n} \}$$
These sets are Lebesgue measurable, hence for every $n \in \mathbb{N}$ and every $i=0 \dots N2^n-1$, there exist Borel sets $E^{i}_n$ and $F^{i}_n$ such that $E^{i}_n \subseteq A^{i}_n \subseteq F^{i}_n$ and $\mathcal{L}(F^{i}_n \setminus E^{i}_n)=0$. Next define two sequences of simple functions $(h_n(x))_n$ and $(g_n(x))_n$ in the following way: $$ g_n(x)=\sum_{i=0}^{N2^n-1} \frac{i}{2^n} \chi_{E^{i}_n} \qquad h_n(x)=\sum_{i=0}^{N2^n-1} \frac{i+1}{2^n} \chi_{F^{i}_n}. $$
Note that $h_n,g_n$ are Borel measurable for every $n$ since they are linear combination of characteristic functions of Borel sets. Finally, set:
$$h(x)=\lim_{n \rightarrow +\infty} h_n(x) \qquad g(x)=\lim_{n \rightarrow +\infty} g_n(x); $$
$h(x)$ and $g(x)$ are Borel measurable since they are pointwise limit of Borel functions. It remains to prove that such limits actually exist for every $x$ (or, as suggested in comments, use the $\limsup$ instead), that the required inequality is satisfied and that $h$ and $g$ coincide a.e.