Prove that every Lebesgue measurable function is equal almost everywhere to a Borel measurable function

You just make sure you have countably many Lebesgue sets in your formulation, and from each of the disjoint sets remove sets of measure $0$ to make them Borel.

Almost everywhere equality follows, and that's all the question cares about.

For any Lebesegue mesurable set $A_i$ exsists a Borel set $C_i$ such that $C_i \subseteq A_i$ and $m(A_i\setminus C_i)=0$. If $A_i$ are disjointed also $C_i$ are disjointed. Is not important that $$\bigcup^{n}_{i=1} {C_i} = \mathbb{R}$$ or, if you prefer, let $C=\bigcup^{n}_{i=1} {C_i}$ then $$ t(x) = \sum^n_{i=i}{\alpha_i\chi_{C_i}(x)}+ 0\chi_{C^c}(x) $$ in the required function ($C^c$ is trivially Borel-measurable).

Hint.

Fact 1. If $f:\mathbb R\to [0,\infty]$ is Lebesgue measurable, then it expressed as $$ f(x)=\sum_{n\in\mathbb N}a_n\chi_{A_n}, $$ where $A_n$ Lebesgue measurable and $a_n\in [0,\infty]$ - This can be shown approximating $f$ from below by simple functions, and then using Lebesgue's Monotone Convergence Theorem.

Fact 2. If $A$ is Lebesgue measurable, then there exists a Borel set $B$ such that $$ m(A\smallsetminus B)+m(B\smallsetminus A)=0. $$