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Im trying to understand a proof for differentiably a.e for functions $F$ given by $$F(x)= \int_{-\infty}^{x}f\ \mathsf dt$$ for $f$ Lebesgue measurable and $L^{1}$. He defines a finite Borel measure $$\mu(A)= \int_{A}f\ \mathsf d \lambda. $$ But then he picks a Borel function $f_0$ equal to $f$ a.e in order to use two lemmas justifying the following equalites $$F'(x)= D \mu(x)=f_0=f\text{ a.e.}$$ The two lemmas are:

  1. If a finite Borel measure is differentiable then its distribution function is differentiable as well.
  2. If $\mu$ is a finite Borel measure then $\mu$ is differentiable w.r.t Lebesgue measure a.e.

I dont understand why he need to pick this $f_0$. For those who have Cohn's Measure Theory this will be Theorem 6.3.6 in latest edition.

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Not a complete solution, but some potentially relevant thoughts.

The Borel function $f_0(x)$ is simply the derivative of $\mu$ at $x$ (where it is differentiable) and $0$ elsewhere (defined by Theorem 6.2.3). Therefore we know that $F'(x)=(D\mu)(x)=f_0(x)$ a.e. (Lemma 6.3.1).

Can you get that $f_0(x)\leq f(x)$ for all $x$ from this? It looks like $f$ is assumed to be non-negative so we just need to show that where $F'(x)$ exists, $F'(x)\leq f(x)$.

Now we just need to show that $f_0 = f$ a.e. You can probably use Proposition 2.2.3 to find an $f_1$ such that $f_0=f_1$ almost everywhere and $f_0\leq f \leq f_1$ everywhere. I think this is the reason why such an $f_0$ is "constructed".

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  • $\begingroup$ ah, so all this because we want a lebegue meaurable function? $\endgroup$ – Jeaj Aug 9 '15 at 6:33
  • $\begingroup$ I think in order to use Prop 2.2.3, we need a function $f_0\leq f$ for all $x$. However, $D\mu$ may not exist everywhere, so we define $f_0=0$ where the derivative $D\mu$ doesn't exist. $f_0$ is definitely a Lebesgue measurable function, but I don't think that's the key point. I think it's relationship with Prop 2.2.3 is important to reconcile. Maybe something with lower and upper derivatives of $\mu$ will complete the argument. $\endgroup$ – jdods Aug 9 '15 at 13:08
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Ive come to the conclusion that this is an unnecessary step. It all follows from uniqness of Radon-Nikodym derivate. But please comment if Im wrong.

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