# Sequences of Borel measurable functions and limit almost everywhere

I've been trying to prove that if $f:D\subset\mathbb{R}\longrightarrow \mathbb{\hat{R}}$ (where $\mathbb{\hat{R}}$ is the extended real number system) is $\textbf{non-negative and measurable}$, then $f$ is the limit almost everywhere of a sequence of Borel measurable functions. I'm starting to believe this is false, but couldn't come up with a counterexample.

Any suggestions? To begin with, is this proposition false as I'm starting to believe? I'm familiar with the concept of limit almost everywhere and Borel measurable functions, but can't seem to connect the two of them.

• Pointwise limit of measurable functions are measurable, and the Lebesgue measure is complete. Hence any function not Lebesgue-measurable serves as a counter-example, e.g., characteristic function of a non-measurable set. – Qiyu Wen Mar 28 '16 at 22:32
• I just noticed I forgot to add that $f$ is measurable. So, since the Lebesgue measure is complete and Borel measurable functions are Lebesgue measurable (but not the other way around), does this mean the proposition is true? Another important consideration is that $f$ is non-negative. – Fawcett512 Mar 28 '16 at 22:39
• In the case that $f$ is measurable, there is a sequence of simple measurable functions $\{s_n\}$ converging to $f$ pointwise. Any book on real analysis probably has such a construction. Technically you can also put $f_n = f$. – Qiyu Wen Mar 28 '16 at 22:43
• And are simple measurable functions Borel measurable? Because in that case I'll have the proposition I've been trying to prove. – Fawcett512 Mar 28 '16 at 22:50
• I also read that every Lebesgue measurable function is equal almost everywhere to a Borel measurable function. So with this result we have the proposition that I'm trying to prove? Seems so since the proposition states the $f$ is the limit almost everywhere of a sequence of Borel measurable functions. – Fawcett512 Mar 28 '16 at 22:53