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Let $\mu$ be a finite Borel measure over $\mathbb{R}$. We define $F(x) = \mu ((-\infty,x])$. Show that $\mu$ is singular with respect to the Lebesgue measure ($m$) if and only if $F'=0$ a.e.

If $F'=0$ a.e., therefore, $F(x) = \mu ((-\infty,x])$ is singular, so in the Lebesgue decomposition its absolutely continuous part is identically $0$. There will be no problem for showing that $\mu$ must be singular with respect to $m$. The complication i'm having is to prove the first implication

ianbounos
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    It would follow from Lebesgue differentiation theorem for finite Borel measures (if you have it available) on $\mathbb{R}$ and the fact that $\mu(\mathbb{R} \setminus A)=0$ for some Lebesgue null set $A$ (as $\mu$ is singular w.r.t. the Lebesgue measure). – Severin Schraven Jul 04 '17 at 11:27

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If $\mu$ is singular w.r.t. the Lebesgue measure $m$, then there exist $A\subseteq \mathbb{R}$ such that

$$ m(A)=0, \qquad \mu(\mathbb{R}\setminus A)=0. $$

Thus, we can write

$$ F(x) = \mu((-\infty, x])= \int_{-\infty}^x 1 d\mu(t) = \int_{-\infty}^x 1_A(t) d\mu(t). $$

By the Lebesgue differentiation theorem for finite Borel measures, we have

$$ F'(x) = 1_A(x) $$

$m-$ a.e. However,

$$ 1_A \equiv 0 $$

$m-$ a.e. (as $m(A)=0$).

You might also want check out the answer of this post: Lebesgue's Theorem on Differentiation of measures?

Severin Schraven
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  • Can you explain why $F'(x)=1_A(x)$? Since $\mu\perp m$, the Radon-Nikodym decomposition of $\mu$ is simply $d\mu=1_Ad\mu+0\cdot dm$. Then by the post you quoted, we have $F'(x)=0$, for $m$-a.e. – Bach Jun 17 '20 at 19:12
  • @Bach As written in my answer above, we have by the Lebesgue differentiation theorem that $F'=1_A$ $m$-a.e. (it is literally the statement of the theorem, I am not sure what I could add to make it more understandable). As $1_A=0$ $m$-a.e. (as $A$ is a null set) we get $F'=0$ $m$-a.e., but all of this I wrote above. Probably I misunderstood your question. – Severin Schraven Jun 18 '20 at 11:37
  • I mean the theorem reads: Let $\nu$ be a regular signed or complex Borel measure on $\mathbb{R}$, and let $d \nu = d \lambda +f dm$ be its Radon-Nikodym representation. Then for $m$-almost every $x \in \mathbb{R}$: $\lim_{r \to 0} \frac{\nu (E_r)}{m(E_r)} = f(x)$ for every family ${E_r}{r>0}$ that shrinks nicely to $x$. Here $f=0$ and $F'(x)=\lim{r \to 0} \frac{\nu (E_r)}{m(E_r)}$. Is that correct? – Bach Jun 18 '20 at 12:01
  • For me it is the following: Let $\nu$ be a finite Borel measure on $\mathbb{R}$ and let $f\in L^1(\mathbb{R})$ and $F(x):=\int_{-\infty}^x f(x) \mu(dx)$. Then $F$ is differentiable $\nu$-a.e. and at those places we have $F'(x)=f(x)$. In the answer above we used $f=1_A$. I looked it up now and saw that this special case is actually called "fundamental theorem of Lebesgue integral calculus". You can use your version with intervals to prove it. – Severin Schraven Jun 18 '20 at 17:25