I read the following theorem in my lecture material stating as follows: For any probability measure $\mu$ on $(\mathbf{R},\,\mathcal{B})$ with a density function $f(x)$, the following condition holds for almost all $x$ $$\lim_{h\downarrow 0}\frac{1}{2h}\mu(\{x-h,\,x+h\})=f(x). $$

My question is, by stating a measure 'with a density', does that mean $\mu\ll Leb$, and $\mu$ does not have singular or discrete part? Besides, what is the name of above theorem, where can I find the proof of it?

  • $\begingroup$ I can only really guess, but: I would also assume that it means that $\mu$ is absolutely continuous w.r.t. the Lebesgue measure such that $\endgroup$ – Furrer Oct 12 '18 at 22:00
  • $\begingroup$ Thanks! But regarding the formula for calculating the density, any clue for its proof? $\endgroup$ – lychtalent Oct 12 '18 at 22:03
  • $\begingroup$ In any case, it will be useful to write the LHS as an integral of $f$ w.r.t. the Lebesgue measure. You could then argue using a link between Lebesgue integration and Riemann integration and using the Fundamental Theorem of Calculus, but that seems a bit too restrictive. Give me a moment. $\endgroup$ – Furrer Oct 12 '18 at 22:15

Assume that $\mu$ is absolutely continuous w.r.t. the Lebesgue measure. This implies the existence of a measurable and Lebesgue integrable function $f$ such that \begin{align*} \mu(-\infty,x] = \int_{-\infty}^x f(y) \, \mathrm{d}y. \end{align*} This shows that the map $F(x) = \mu(-\infty,x]$ is absolutely continuous. (The classic link between probability densities and Radon-Nikodym derivatives.) In particular, $F$ is almost everywhere differentiable with derivative $f$. (I suppose this result is known to you. Otherwise see Royden's or Rudin's.) Can you take it from here?

  • $\begingroup$ If I understand correctly, shouldn't it be $\lim_{h\downarrow 0}\frac{1}{2h}\mu((x-h,\,x+h])=f(x) $ rather than $\lim_{h\downarrow 0}\frac{1}{2h}\mu(\{x-h,\,x+h\})=f(x)$? $\endgroup$ – lychtalent Oct 12 '18 at 22:49
  • $\begingroup$ The set $\{x-h\}$ is a Lebesgue null-set and hence also a $\mu$ null-set. Thus the LHS's are identical. But anyway, I would prefer the first expression (it has natural connotations to how we define the Lebesgue measure on sets $(a,b]$). $\endgroup$ – Furrer Oct 12 '18 at 22:53
  • $\begingroup$ @Furrer You are absolutely right. $\mu (\{x-h,x+h\})=0$ so the formula is wrong. $\endgroup$ – Kavi Rama Murthy Oct 12 '18 at 23:14
  • $\begingroup$ When lychtalent writes $\{x-h,x+h\}$, I think he means the closed set which I denote $[x-h,x+h]$, so the formula is right. For a singleton $x$ I write $\{x\}$. $\endgroup$ – Furrer Oct 12 '18 at 23:25
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    $\begingroup$ You are right. Let me also point out that when $\mu << m$ it does not matter whether the interval from $x-h$ to $x+h$ includes the end points or not. It can be open, closed or half open. $\endgroup$ – Kavi Rama Murthy Oct 12 '18 at 23:28

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