This is a verification request for a counterexample that I think I have found for Problem 32.13 on page 427 in Patrick Billingsley’s Probability and Measure textbook (third edition, but the problem appears unaltered in other editions, too).

The result to be proven in this problem is quoted (almost) verbatim:

Suppose that $\mu$ is a Borel probability measure on $\mathbb R$, that $\nu$ is a $\sigma$-finite Borel measure on $\mathbb R$, and that $\nu\ll\mu$ (that is, $\nu$ is absolutely continuous with respect to $\mu$). Show that the Radon–Nikodym derivative $f$ satisfies $$\lim_{h\downarrow0}\frac{\nu(x-h,x+h]}{\mu(x-h,x+h]}=f(x)$$ on a set of $\mu$-measure $1$.

I can prove this result if $\nu$ is assumed to be finite (using an earlier result presented in Problem 31.22 on page 419; the key condition is the $\mu$-integrability of the Radon–Nikodym derivative), or even regular (that is, finite on compact sets—thanks to Dominique R.F. for pointing this out in a comment below). However, the assumption that $\nu$ is merely $\sigma$-finite (but potentially irregular) does not seem to be strong enough to ensure that the current result holds. I present a counterexample below, for which I welcome any feedback.

Let $p$ denote the point mass at the point $1/2$. This is a probability measure that assigns each Borel subset $A$ of the line a value $p(A)\in\{0,1\}$ according as $1/2\in A$ or $1/2\notin A$. Furthermore, let $\lambda$ denote the Lebesgue measure. Define, for each Borel set $A$, $$\mu(A)\equiv\frac{1}{2}\lambda(A\cap(0,1))+\frac{1}{2}p(A).$$ It is easy to check that $\mu$ is a probability measure on $\mathbb R$ and is supported on $(0,1)$. Furthermore, if $f:\mathbb R\to[0,\infty)$ is a real-valued, non-negative, Borel-measurable function, then it is not difficult to prove that $$\int f(t)\,\mathrm d\mu(t)=\frac{1}{2}\int_0^1f(t)\,\mathrm d t+\frac{1}{2}f\left(\frac{1}{2}\right),$$ where $\mathrm dt$ denotes simply integration with respect to the Lebesgue measure.

Now consider the following function on $\mathbb R$: \begin{align*} f(x)\equiv\begin{cases}\dfrac{1}{x-\dfrac{1}{2}}&\text{if $x>\dfrac{1}{2}$,}\\\dfrac{1}{\dfrac{1}{2}-x}&\text{if $x<\dfrac{1}{2}$,}\\0&\text{if $x=\dfrac{1}{2}$.}\end{cases} \end{align*} Clearly, $f$ is real-valued, non-negative, and Borel-measurable.

Now define, for each Borel set $A$, $$\nu(A)\equiv\int_Af(t)\,\mathrm d\mu(t).$$ Then, $\nu$ is a measure and it is $\sigma$-finite, because letting $$A_n\equiv\{x\in\mathbb R\,|\,n-1\leq f(x)<n\}\quad\text{for each $n\in\mathbb N$},$$ it is easily seen that $$\mathbb R=\bigcup_{n\in\mathbb N} A_n$$ (given that $f$ never assumes infinite or negative values) and that $$\nu(A_n)=\int_{A_n}f(t)\,\mathrm d\mu(t)\leq\int_{A_n}n\,\mathrm d\mu(t)=n\mu(A_n)\leq n<\infty$$ for each $n\in\mathbb N$. Furthermore, $\mu(A)=0$ implies that $\nu(A)=0$ for any Borel set $A$, and $f$ is the Radon–Nikodym derivative of $\nu$ with respect to $\mu$ (almost unique with respect to $\mu$).

Therefore, all the premises corresponding to Billingsley’s statement are satisfied. Letting $x=1/2$ and $h\in(0,1/2)$, one can see that $$\nu\left(\frac{1}{2}-h,\frac{1}{2}+h\right]=\int_{\left(\frac{1}{2}-h,\frac{1}{2}+h\right]}f(t)\,\mathrm d\mu(t)=\frac{1}{2}\int_{\frac{1}{2}-h}^{\frac{1}{2}+h}f(t)\,\mathrm dt+\frac{1}{2}\underbrace{f\left(\frac{1}{2}\right)}_{=0}=\infty,$$ given how $f$ has been defined. On the other hand, $\mu\left(\frac{1}{2}-h,\frac{1}{2}+h\right]=h+\frac{1}{2}>0,$ so that $$\lim_{h\downarrow0}\dfrac{\nu\left(\dfrac{1}{2}-h,\dfrac{1}{2}+h\right]}{\mu\left(\dfrac{1}{2}-h,\dfrac{1}{2}+h\right]}=\infty\neq0=f\left(\frac{1}{2}\right).$$ Hence, the conclusion does not hold at $x=1/2$, so it cannot hold almost everywhere with respect to $\mu$ given that $\mu(\{1/2\})=1/2>0$.

Clearly, $\nu$ in this counterexample is irregular; even though it is $\sigma$-finite, an infinite mass “accumulates” in arbitrarily small neighborhoods around the point $1/2$.

  • 2
    $\begingroup$ To corroborate your doubts, the corresponding theorem in Bogachev's Measure Theory (theorem 5.8.8) has the stronger hypothesis that $\nu$ is finite on all balls. This seems to be exactly what goes wrong in your example. $\endgroup$ Oct 15, 2016 at 23:28
  • $\begingroup$ @DominiqueR.F. I see! So the key condition here seems to be regularity as opposed to $\sigma$-finitude (of course, regularity is a stronger condition than $\sigma$-finitude and is weaker than finitude). $\endgroup$
    – triple_sec
    Oct 15, 2016 at 23:50


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