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I was reading this old question:

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

I've found out a solution by myself,but I can't undestand an answer that an user gave.

The answer is in 2 steps. The first step claims that every non-negative lebesgue function can be written as countable sum of terms $\chi_{A_n} \alpha_n$, with $A_n$ lebesgue measurable set. I think that's false, it looks strange. Can anyone help me to clarify this?

Naj Kamp
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2 Answers2

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This is just a reformulation of something you find in any book on measure theory:

If $f\ge0$ is measurable then there is a sequence of non-negative simple functions $\phi_n$ increasing to $f$.

Indeed, saying $\phi_n\to f$ is the same as $$f=\phi_1+(\phi_2-\phi_1)+\dots.$$

For each fixed $n$, the function $\phi_{n+1}-\phi_n$ is a non-negative simple function, so it is a finite sum $$\phi_{n+1}-\phi_n=\sum\alpha_j\chi_{E_j},$$with $\alpha_j\ge0$.

David C. Ullrich
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  • I see thanks. But if that's true, then $Im(f)$ is countable . Or I'm missing something? – Naj Kamp Dec 23 '17 at 15:45
  • Yes, you're missing something - it certainly does not follow that the range is countable. Hard to say exactly what you're missing unless you tell us why you think it does follow. (This is where you "accept" the answer by clicking on that little checkmark, by the way...) – David C. Ullrich Dec 23 '17 at 15:50
  • Ah, maybe this is what you're missing: I said "For each fixed $n$". For each $n$ we can take the $E_j$ to be disjoint. But the $E_j$ for one $n$ are not disjoint from the $E_j$ for another $n$, so the range of $f$ is not given by the set of all the $\alpha_j$. – David C. Ullrich Dec 23 '17 at 15:52
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Let me show a slightly different approach, that uses base $2$. For simplicity, I will only consider $f$ with $0\leq f\leq1$. This is no hurdle, since $$ f=\sum_{n=1}^\infty [n+f_n]\,1_{\{n-1\leq f\leq n\}}, $$ where $f_n=f-n$ satisfies $0\leq f_n\leq 1$.

For $0\leq f\leq 1$, then we have $$ f=\sum_{n=1}^\infty 2^{-n}\,1_{\{2^nf\in E_n\}}^\vphantom{E}=\sum_{n=1}^\infty 2^{-n}\,1_{f^{-1}(2^{-n}E_n)}, $$ where $$ E_n=\bigcup_{n=1}^\infty [2n-1,2n) $$ Basically, the sets $E_n$ are used to detect if the $n^{\rm th}$ binary digit is 1 (for that, we use the "odd" intervals).

Martin Argerami
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