Lebesgue-measurable almost everywhere equals Borel measurable function

Question

Let $\, f: \mathbb{R}^n \rightarrow [-\infty, +\infty]$ be a Lebesgue measurable function.

I want to show that f is $\lambda_n$- alomost everywhere equal to to a Borel measurable function $\, f'$.

I would very much appreciate your help.

Best, KingDingeling

Answer 1

I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of $\{\infty\}$, $\{-\infty\}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.

Answer 2

Sketch (just an elaboration of the other answer):

• $f$ is the pointwise limit of simple functions $f_n.$
• Each $f_n$ is a sum of terms $c\chi_A$ where $c\in \mathbb R$ and $A$ is Lebesgue measurable.
• Each $A=E\cup N$ where $E$ is a Borel set and $\lambda (N)=0.$
• Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show
• $f$ is the pointwise limit of $g_n$.