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I've been given a homework exercise that has been troubling me, and unless I get some help I'll go crazy and not get anything else done this weekend. The exercise is stated (more clearly than the title) as follows:

Let $\{\mu_n\}_{n=1}^{\infty}$ be a sequence of Borel measures on $\mathbb{R}$ so that each $\mu_n$ is finite on bounded Borel sets. Prove that there is a Borel measure that is finite on bounded sets and such that a Borel set $E\in\mathcal{B}_{\mathbb{R}}$ is of measure zero for $\mu$ if and only if $\mu_n(E)=0$ for each $n\in\mathbb{N}$.

Along with the exercise, the professor of our course gives a hint to try defining $\mu$ as $\sum_{n=1}^{\infty}2^{-n}\mu_n(A)$ and goes on to say that this definition does not work because given a bounded $E\in\mathcal{B}_{\mathbb{R}}$ we are not guaranteed that $\sum_{n=1}^{\infty}2^{-n}\mu_n(E)<\infty$.

But since this was given a hint, I assume that with a slight modification the idea could work.

Attempts

  • I thought about defining $\mu$ has $\sum_{n=1}^{\infty}2^{-n}f\left(\mu_n(A)\right)$, where $f$ is a bounded function from the nonnegative reals to itself such that $f(x+y)=f(x)+f(y)$ and $f(0)=0$, however I don't believe such a function exists. (Such a function would satisfy $f(n)=nf(1)$, so $f$ can't be bounded)
  • My next thought was maybe I could find some sequence $\{a_n\}_{n=1}^{\infty}$ which converges to zero so rapidly as to force convergence of $\sum_{n=1}^{\infty}a_n\mu_n(A)$ for all Borel sets, but this shouldn't work either because given a sequence that converges to zero, there's no guarantee that there does not exist a Borel set E so that $\{\mu_n(E)\}_{n=1}^{\infty}$ converges to infinity at a faster rate.
  • Next, after a couple hours of thinking and flipping through my notes, I noticed we can identify each $\mu_n$ with a right continuous increasing function $f_n$ (where $\mu_n\big([a,b)\big)=f(b)-f(a)$ on intervals), so given a bounded interval $[a,b)$ the attempted definition would become $\mu\big([a,b)\big)=\sum_{n=1}^{\infty}f_n(b)-f_n(a)$, but then this still wouldn't work because the family $\{f_n\}_{n=1}^{\infty}$ could be something like $f_n(x)=0$ for $x<0$ and $f_n(x)=n$ for $x\geq 0$; then $\mu_n\big([-1,1)\big)=n$ on the bounded set $[-1,1)$. This puts me in the same situation as the first bullet point.
  • I also had a vague idea of taking the family of bounded Borel sets $A$ where $\sum_{n=1}^{\infty}2^{-n}\mu_n(A)<\infty$ and possibly showing that this is an algebra. Then I could use this premeasure to generate the needed measure. I haven't thought this idea all the way through however.

Any hints or general help would be greatly appreciated. Please no full solutions as this is a homework exercise and I don't need the guilt. Thanks

Edit: To the person who commented: the problem was meant to say "is of measure zero." Like a moron I accidentally edited the comment instead of replying, so I just ended up deleting it.

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    $\begingroup$ Since you can associate each $\mu_n$ with a increasing, right continuous function $f_n$, can you define $$f = \sum_{n=1}^\infty 2^{-n} \tan^{-1} \circ f_n$$ and let $\mu$ be the Lebesgue-Stieltjes measure associated with $f$? $\endgroup$ – Henricus V. Oct 9 '16 at 1:17
  • $\begingroup$ That idea definitely has potential! Thanks! It's funny because (as the first bullet point suggests) this was one of my very first ideas, but I was trying $$ A\mapsto\sum_{n=1}^{\infty}2^{-n}\tan^{-1}\big(\mu_n(A)\big) $$ instead of what you've suggested. Then later I thought about messing with the increasing, right continuous functions $f_n$. I can't believe I haven't connected the dots and tried this. $\endgroup$ – Blake Oct 9 '16 at 1:21
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Hint: Try to first define, for each interval $I_n = [n, n+1)$, a measure $\nu_n$ defined on Borel subsets of $I_n$ such that for every Borel $B \subseteq I_n$, $\nu_n(B) = 0$ iff for every $k$, $\mu_k(B) = 0$. Then paste together these measure.

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  • $\begingroup$ I see. So for each $I_n=[n,n+1)$, for any Borel $B\subset I_n$, the sequence $\{\mu_k(B)\}_{k=1}^{\infty}$ can only diverge as fast as $\{\mu_k(I_n)\}_{k=1}^{\infty}$, so defining $$ \nu_n(B)=\sum_{\big\{k\in\mathbb{N}\,:\,\mu_k(I_n)\neq 0\big\}}2^{-k}\frac{\mu_{k}(B)}{\mu_k(I_n)} $$ appears to do the trick. Very clever! $\endgroup$ – Blake Oct 10 '16 at 9:06

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