I have the following problem: Show $\beta \ll \eta$ if and only if for every $\epsilon > 0 $ there exists a $\delta>0$ such that $\eta(E)<\delta$ implies $\beta(E)<\epsilon$.

For the forward direction I had a proof, but it relied on the use of the false statement that "$h$ integrable implies that $h$ is bounded except on a set of measure zero".

I had no problem with the backward direction.

  • $\begingroup$ Then what exactly is the question? $\beta << \eta \Rightarrow \varepsilon,\delta$ statement? And the definition of absolutely continuos is $\beta = f\eta$ with $\eta$-integrable $f$? If yes, use characteristic functions of measurable sets. $\endgroup$ – user20266 Aug 19 '12 at 15:07
  • $\begingroup$ The question is how to do the forward direction, so yes. $\endgroup$ – Frank White Aug 19 '12 at 15:08
  • $\begingroup$ And what exactly do you mean with the characteristic functions? $\endgroup$ – Frank White Aug 19 '12 at 15:11
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    $\begingroup$ To do the forward direction you need some restriction on $\beta$. The following example is from Rudin's "Real & Complex Analysis": Take $X=(0,1)$, $\eta$ the Lebesgue measure and $\beta(E) = \int_E \frac{1}{t} dt$. Then $\beta \ll \eta$, but the $\epsilon-\delta$ characterization doesn't hold. $\endgroup$ – copper.hat Aug 19 '12 at 16:52
  • $\begingroup$ What if our set is of finite measure? $\endgroup$ – Frank White Aug 19 '12 at 16:52

Assume that $\beta=h\eta$ with $h\geqslant0$ integrable with respect to $\eta$, in particular $\beta$ is a finite measure. Let $\varepsilon\gt0$.

There exists some finite $t_\varepsilon$ such that $\beta(B_\varepsilon)=\int_{B_\varepsilon} h\,\mathrm d\eta\leqslant\varepsilon$ where $B_\varepsilon=[h\geqslant t_\varepsilon]$. Note that, for every measurable $A$, $A\subset B_\varepsilon\cup(A\setminus B_\varepsilon)$, hence $\beta(A)\leqslant\beta(B_\varepsilon)+\beta(A\cap[h\leqslant t_\varepsilon])\leqslant\varepsilon+t_\varepsilon\eta(A)$.

Let $\delta=\varepsilon/t_\varepsilon$. One sees that, for every measurable $A$, if $\eta(A)\leqslant\delta$, then $\beta(A)\leqslant2\varepsilon$, QED.

  • $\begingroup$ For the line where "$\beta(A) \leq \beta(h \geq t_{\epsilon})...$, could you explain this a bit more? I'm having problems seeing where that comes from. $\endgroup$ – Frank White Aug 19 '12 at 15:25
  • $\begingroup$ So, for any measurable set $A$, and given an $\epsilon>0$ we have that $\beta(A)=\int_Ahd\eta$ integrable implies that we can find a $t_{\epsilon}$ such that $\beta(A_{\epsilon})\leq \frac{\epsilon}{2}$, with $A_{\epsilon}=\lbrace x\in A| h(x) \geq t_{\epsilon} \rbrace$. Then $A_{\epsilon}^C=\lbrace x\in A| h(x) < t_{\epsilon} \rbrace$. So, taking $\delta<\frac{\epsilon}{2t_\epsilon}$ we have that $\beta(A_{\epsilon})\leq \beta(A_{\epsilon}) + \beta(A_{\epsilon}^C) \leq \frac{\epsilon}{2} + t_{\epsilon}\eta(A_{\epsilon}^C$). $\endgroup$ – Frank White Aug 19 '12 at 16:30
  • $\begingroup$ Now, if $\eta(A_{\epsilon})< \delta$ then we have from above that $\eta(A_{\epsilon}^C)\leq \eta(A_{\epsilon})< \delta$, thus $\beta(A_{\epsilon}) \leq \frac{\epsilon}{2}+t_{\epsilon}\frac{\epsilon}{2t_{\epsilon}}=\epsilon$. Is this correct? $\endgroup$ – Frank White Aug 19 '12 at 16:33
  • $\begingroup$ Note that this answer assumes that $\beta$ is finite. This is a reasonable assumption, but it is not immediately obvious. $\endgroup$ – copper.hat Aug 19 '12 at 17:13
  • $\begingroup$ @copper.hat Indeed, requiring $h$ to be integrable is assuming that $\beta$ is finite and I could have made this more obvious (thanks). $\endgroup$ – Did Aug 19 '12 at 19:14

No need to use Radon-Nikodým here. I'll assume that $\beta$ is totally finite. See the end of the answer why this is necessary.

Suppose that $\beta E = 0$ whenever $E$ is measurable and $\eta E = 0$ but that the desired $\varepsilon$-$\delta$-condition doesn't hold.

Then there is $\varepsilon \gt 0$ such that for all $\delta \gt 0$ there is $E$ such that $\eta(E) \lt \delta$ but $\beta(E) \geq \varepsilon$. For each $n$ choose $E_n$ such that $\eta(E_n) \lt 2^{-n}$ and $\beta(E_n) \geq \varepsilon$.

Define $E = \bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_n$. Then $$ 0 \leq \eta E \leq \inf_{N \in \mathbb N} \eta \bigcup_{n \geq N} E_n \leq \inf_{N \in \mathbb{N}} \sum_{n \geq N} 2^{-n} = 0 $$ hence $\eta E = 0$. By hypothesis it follows that $\beta E = 0$ as well. On the other hand, assuming $\beta$ is totally finite, we get $$ 0 = \beta E = \lim_{N \to \infty} \beta \bigcup_{n \geq N} E_n \geq \varepsilon \gt 0, $$ which is absurd.

Note that, as @copper.hat pointed out in the comments, it is necessary to assume that $\beta$ is totally finite. The $\sigma$-finite example $\beta E = \int_{E} \frac{1}{t}\,dt$ on $(0,1)$ shows this: For Lebesgue measure $\lambda$ on $(0,1)$, the absolute continuity condition “$\lambda E = 0$ implies that $\beta E = 0$” holds, while the $\varepsilon$-$\delta$-characterization doesn't. For every $\delta \gt 0$ we have $\beta(0,\delta) = \infty$ while $\lambda(0,\delta) = \delta$.

  • $\begingroup$ That $\eta(E) = 0$ follows also from Borel-Cantelli lemma. (actually, one can easily prove this lemma using your argument) $\endgroup$ – Ahriman Aug 19 '12 at 17:33
  • $\begingroup$ That's right, but I tried to keep this argument as elementary as possible. Thanks! $\endgroup$ – t.b. Aug 19 '12 at 17:36

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